M7 Kinematics — revision question pack

5 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section M7. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

M7.1 · Understand and use the language of kinematics: position; displacement; distance travelled; velocity; speed; acceleration.

Explanation

Position locates a particle relative to an origin, displacement is the signed change in position, and distance travelled is the total path length and is always non-negative.
Velocity is the rate of change of displacement, speed is the magnitude of velocity, and acceleration is the rate of change of velocity; choose a positive direction before assigning signs.
For motion from $x_1$ to $x_2$ , displacement is $x_2-x_1$ ; split reversals into separate path lengths for distance, and use change in velocity divided by time for average acceleration.
A common error is to treat distance and displacement, or speed and velocity, as interchangeable; distance and speed cannot be negative.

Worked example

Taking east as positive, a cyclist's velocity changes uniformly from $+8\,\text{m s}^{-1}$ to $-4\,\text{m s}^{-1}$ in $6\,\text{s}$ . Find the acceleration, the time when the cyclist is instantaneously at rest, and the speed at the end.

1.The acceleration is the change in velocity divided by time: $a=(-4-8)/6=-2\,\text{m s}^{-2}$ .
2.Using $v=u+at$ , rest occurs when $0=8-2t$ , so $t=4\,\text{s}$ .
3.The final velocity is $-4\,\text{m s}^{-1}$ , whose magnitude gives speed $4\,\text{m s}^{-1}$ .

Answer: Acceleration $=-2\,\text{m s}^{-2}$ ; The cyclist is at rest after $4\,\text{s}$ ; Final speed $=4\,\text{m s}^{-1}$

Common mistakes

Don't calculate displacement by adding journey lengths without assigning directions.
Don't report a negative velocity as a negative speed and ignore the chosen positive direction.

Exam tip

Keep signs for displacement, velocity and acceleration, but report distance and speed as non-negative magnitudes.

Tier 1 · Easy

1.
A particle moves on a straight line from position $-3\,\text{m}$ to $5\,\text{m}$ and then to $1\,\text{m}$ . Find its displacement and the distance it travels.
(3)
(Total for Question 1 is 3 marks)
2.
Taking upwards as positive, a lift has velocity $-3.5\,\text{m s}^{-1}$ . State its speed and direction of motion.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
At one instant a particle moving on a straight line has velocity $-7\,\text{m s}^{-1}$ and acceleration $+2\,\text{m s}^{-2}$ . State its speed and explain whether its speed is increasing or decreasing at that instant.
(3)
(Total for Question 1 is 3 marks)
2.
Taking right as positive, the velocity of a particle changes uniformly from $-6\,\text{m s}^{-1}$ to $2\,\text{m s}^{-1}$ in $4\,\text{s}$ . Find its acceleration. State whether its speed is initially increasing or decreasing, and whether the particle changes direction during the interval.
(4)
(Total for Question 2 is 4 marks)
3.
Taking right as positive, a particle has velocity $-6\,\text{m s}^{-1}$ at time $t=0$ . During the next $2\,\text{s}$ its displacement is $-4\,\text{m}$ . State its speed at $t=0$ and find its average velocity during the interval. Explain why the information given is not sufficient to find its average speed during the interval.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A runner travels $120\,\text{m}$ east in $15\,\text{s}$ , $50\,\text{m}$ west in $5\,\text{s}$ and then $30\,\text{m}$ east in $10\,\text{s}$ . Find the runner's total distance, displacement, average speed and average velocity.
(6)
(Total for Question 1 is 6 marks)
2.
A particle starts at position $10\,\text{m}$ on a directed straight line. It moves with constant velocity $-3\,\text{m s}^{-1}$ for $4\,\text{s}$ , then with velocity $5\,\text{m s}^{-1}$ for $6\,\text{s}$ , and finally with velocity $-2\,\text{m s}^{-1}$ for $3\,\text{s}$ . Find its final position, total distance travelled, average speed and average velocity over the complete motion.
(6)
(Total for Question 2 is 6 marks)
3.
A particle starts at position $-5\,\text{m}$ on a directed straight line and initially moves in the positive direction. During $8\,\text{s}$ it changes direction exactly once and finishes at position $7\,\text{m}$ . Its average speed during the interval is $3.5\,\text{m s}^{-1}$ . Find the distance it travels in each direction, the position at which it changes direction and its average velocity over the interval.
(6)
(Total for Question 3 is 6 marks)
4.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A survey robot starts at position $(-2\mathbf{i}+\mathbf{j})\,\text{m}$ and finishes at $(16\mathbf{i}+25\mathbf{j})\,\text{m}$ . Its odometer records a total distance of $42\,\text{m}$ during the $14\,\text{s}$ journey. Find its displacement vector and the magnitude of its displacement. Hence find its average speed and its average velocity vector, and explain why the magnitude of the average velocity is less than the average speed. Give all numerical answers exactly.
(6)
(Total for Question 4 is 6 marks)
5.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. During a $6\,\text{s}$ interval, a particle moves from $(-4\mathbf{i}+3\mathbf{j})\,\text{m}$ to $(20\mathbf{i}-9\mathbf{j})\,\text{m}$ . Its velocity changes from $(-3\mathbf{i}+4\mathbf{j})\,\text{m s}^{-1}$ to $(5\mathbf{i}-8\mathbf{j})\,\text{m s}^{-1}$ . Find the particle's average velocity and average acceleration during the interval, and its speed at each endpoint. Explain why these data are insufficient to determine the total distance travelled. Give all numerical answers exactly.
(6)
(Total for Question 5 is 6 marks)

M7.2 · Understand, use and interpret graphs in kinematics for motion in a straight line: displacement against time and interpretation of gradient; velocity against time and interpretation of gradient and area under the graph.

Explanation

The gradient of a displacement-time graph is velocity; the gradient of a velocity-time graph is acceleration, with signs determined by the chosen positive direction.
Find displacement from a velocity-time graph using signed area, splitting the graph into rectangles, triangles or trapezia as needed.
For a straight segment from velocity $u$ to $v$ over time $t$ , the area is $\dfrac12(u+v)t$ , which gives the displacement during that interval.
A common error is to add all areas as positive: area below the time axis is negative for displacement, though its magnitude contributes positively to distance travelled.

On a velocity-time graph, gradient gives acceleration and signed area gives displacement.

Worked example

A particle's velocity increases uniformly from $2\,\text{m s}^{-1}$ to $8\,\text{m s}^{-1}$ during the first $3\,\text{s}$ , then remains at $8\,\text{m s}^{-1}$ for $4\,\text{s}$ . Find its acceleration during the first stage and its displacement over all $7\,\text{s}$ .

1.The first gradient is $(8-2)/3=2\,\text{m s}^{-2}$ .
2.The first-stage area is $\tfrac12(2+8)(3)=15\,\text{m}$ and the constant-velocity area is $8(4)=32\,\text{m}$ , giving displacement $15+32=47\,\text{m}$ .

Answer: Acceleration $=2\,\text{m s}^{-2}$ ; Displacement $=47\,\text{m}$

Common mistakes

Don't treat the area under a displacement-time graph as displacement instead of using its gradient for velocity.
Don't read displacement from the final velocity instead of the signed area under the velocity-time graph.

Exam tip

On kinematics graphs, state explicitly whether a gradient or signed area gives the requested quantity.

Tier 1 · Easy

1.
A straight segment of a displacement-time graph joins $(2,5)$ to $(8,23)$ , where time is in seconds and displacement in metres. Find the velocity represented by the segment.
(2)
(Total for Question 1 is 2 marks)
2.
A particle starts from rest and its velocity increases linearly to $7\,\text{m s}^{-1}$ after $4\,\text{s}$ . Find the displacement during these $4\,\text{s}$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A displacement-time graph consists of straight line segments joining $(0,0)$ to $(4,20)$ and then $(4,20)$ to $(10,8)$ , with time in seconds and displacement in metres. Find the velocity on each segment and the total distance travelled.
(4)
(Total for Question 1 is 4 marks)
2.
A particle has velocity $3\,\text{m s}^{-1}$ at $t=0$ . Its velocity increases linearly to $V\,\text{m s}^{-1}$ at $t=4$ , then decreases linearly to $1\,\text{m s}^{-1}$ at $t=10$ . The displacement from $t=0$ to $t=10$ is $34\,\text{m}$ . Find $V$ and the acceleration during each stage.
(5)
(Total for Question 2 is 5 marks)
3.
At $t=4\,\text{s}$ , the tangent to a particle's displacement-time graph passes through the points $(2,11)$ and $(7,-4)$ , where displacement is measured in metres. Find the particle's velocity at $t=4\,\text{s}$ . At $t=9\,\text{s}$ the graph has a horizontal tangent. State what this shows about the particle's velocity at that instant.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A particle has velocity $4\,\text{m s}^{-1}$ at $t=0$ . Its velocity increases linearly to $10\,\text{m s}^{-1}$ at $t=3$ , remains constant until $t=5$ , then decreases linearly to $-2\,\text{m s}^{-1}$ at $t=9$ . Find the acceleration during the final stage, the displacement and the total distance travelled from $t=0$ to $t=9$ .
(7)
(Total for Question 1 is 7 marks)
2.
A particle's velocity decreases uniformly from $8\,\text{m s}^{-1}$ at $t=0$ to a negative value at $t=T$ . During this interval its displacement is $20\,\text{m}$ and its total distance travelled is $\dfrac{100}{3}\,\text{m}$ . Find $T$ , the final velocity and the time at which the particle changes direction.
(6)
(Total for Question 2 is 6 marks)
3.
A velocity-time graph begins with a straight line segment from $(0,2)$ to $(4,0)$ , where time is in seconds and velocity in metres per second. For $4<t<10$ , the velocity is modelled by the semicircular arc lying below the time axis, with endpoints $(4,0)$ and $(10,0)$ and minimum velocity $-3\,\text{m s}^{-1}$ ; the model does not specify the acceleration at $t=4$ or $t=10$ . Determine the constant acceleration represented by the straight segment, the exact displacement from $t=0$ to $t=10$ , and the exact total distance travelled during this interval.
(6)
(Total for Question 3 is 6 marks)
4.
A displacement-time graph consists of straight line segments joining $(0,2)$ to $(3,k)$ and then $(3,k)$ to $(8,-4)$ , where time is in seconds and displacement in metres. The particle travels a total distance of $30\,\text{m}$ and $k>2$ . Find $k$ , the velocity represented by each segment, the average speed and the average velocity over the complete motion. Give all numerical answers exactly.
(6)
(Total for Question 4 is 6 marks)
5.
Particles $A$ and $B$ move on the same directed straight line. At $t=0$ , $A$ is at the origin and moves with constant velocity $6\,\text{m s}^{-1}$ , while $B$ is $6\,\text{m}$ ahead. The velocity-time graph for $B$ is the straight line through $(0,2)$ with gradient $1\,\text{m s}^{-2}$ . Using areas under the velocity-time graphs, find all positive times at which the particles are at the same position. For each meeting, find the common position and state which particle is moving faster immediately after the meeting. Give all numerical answers exactly.
(7)
(Total for Question 5 is 7 marks)

M7.3 · Understand, use and derive the formulae for constant acceleration for motion in a straight line; extend to 2 dimensions using vectors.

Explanation

For constant acceleration, use $\mathbf{v}=\mathbf{u}+\mathbf{a}t$ and $\mathbf{r}=\mathbf{r}_0+\mathbf{u}t+\dfrac12\mathbf{a}t^2$ , with scalar forms available for one-dimensional motion. List the known quantities from $s,u,v,a,t$ , choose an equation containing the required unknown, and keep the sign convention consistent throughout.
Eliminating $t$ between $v=u+at$ and $s=\dfrac12(u+v)t$ gives $v^2=u^2+2as$ for motion with constant acceleration.
A common error is to use constant-acceleration formulae when acceleration varies, or to use vector magnitudes before completing the component equations. $v=u+at=4+1.5(6)=13\,\text{m s}^{-1}$ .
Also $s=ut+\tfrac12at^2=4(6)+\tfrac12(1.5)(6^2)=24+27=51\,\text{m}$ .
Confirm that the time is within the stated motion interval.

Worked example

A car moves in a straight line with initial speed $4\,\text{m s}^{-1}$ and constant acceleration $1.5\,\text{m s}^{-2}$ . Find its speed and the distance it travels in the next $6\,\text{s}$ .

1. $v=u+at=4+1.5(6)=13\,\text{m s}^{-1}$ .
2.Also $s=ut+\tfrac12at^2=4(6)+\tfrac12(1.5)(6^2)=24+27=51\,\text{m}$ .

Answer: Speed $=13\,\text{m s}^{-1}$ ; Distance $=51\,\text{m}$

Common mistakes

Don't mix a displacement sign convention with an acceleration value treated as an unsigned magnitude.
Don't select a constant-acceleration formula containing an extra unknown and introduce avoidable algebra.

Exam tip

List the known kinematic quantities with signs, then choose the equation containing only the required unknown.

Tier 1 · Easy

1.
A particle accelerates uniformly from $5\,\text{m s}^{-1}$ to $11\,\text{m s}^{-1}$ in $3\,\text{s}$ . Find its acceleration.
(2)
(Total for Question 1 is 2 marks)
2.
A train travels with initial speed $9\,\text{m s}^{-1}$ and constant acceleration $1.2\,\text{m s}^{-2}$ . Find the distance it travels in $5\,\text{s}$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A particle has initial velocity $(3\mathbf{i}+4\mathbf{j})\,\text{m s}^{-1}$ and constant acceleration $(2\mathbf{i}-\mathbf{j})\,\text{m s}^{-2}$ . Find its velocity and displacement after $5\,\text{s}$ .
(4)
(Total for Question 1 is 4 marks)
2.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A particle moves with constant acceleration $(2\mathbf{i}+3\mathbf{j})\,\text{m s}^{-2}$ . Its displacement during the first $4\,\text{s}$ is $(24\mathbf{i}+4\mathbf{j})\,\text{m}$ . Find its initial velocity and its velocity at $t=4$ .
(4)
(Total for Question 2 is 4 marks)
3.
For motion with constant acceleration, eliminate $t$ from $v=u+at$ and $s=\tfrac12(u+v)t$ to derive $v^2=u^2+2as$ . Hence find the distance required for a particle moving at $13\,\text{m s}^{-1}$ to come to rest with constant acceleration $-2.6\,\text{m s}^{-2}$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A particle starts at the origin with velocity $(3\mathbf{i}+8\mathbf{j})\,\text{m s}^{-1}$ and constant acceleration $(2\mathbf{i}-2\mathbf{j})\,\text{m s}^{-2}$ . Find the time when its velocity is parallel to $\mathbf{i}$ , its displacement then, and its speed then.
(6)
(Total for Question 1 is 6 marks)
2.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A particle starts from the origin with velocity $(4\mathbf{i}+\mathbf{j})\,\text{m s}^{-1}$ . For the first $3\,\text{s}$ its acceleration is $(\mathbf{i}-2\mathbf{j})\,\text{m s}^{-2}$ . Its acceleration then changes instantaneously to $(-2\mathbf{i}+\mathbf{j})\,\text{m s}^{-2}$ for a further $2\,\text{s}$ . Find its velocity and position vector at the end of the $5\,\text{s}$ .
(6)
(Total for Question 2 is 6 marks)
3.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A particle starts at the origin with constant acceleration. Its position vectors after $2\,\text{s}$ and $5\,\text{s}$ are $(10\mathbf{i}+6\mathbf{j})\,\text{m}$ and $(40\mathbf{i}+45\mathbf{j})\,\text{m}$ respectively. Find its initial velocity, its acceleration and its velocity after $5\,\text{s}$ .
(6)
(Total for Question 3 is 6 marks)
4.
Particle $A$ passes a point $O$ at time $t=0$ and continues in a straight line with constant speed $10\,\text{m s}^{-1}$ . At $t=2\,\text{s}$ , particle $B$ starts from rest at $O$ and moves along the same line with constant acceleration $4\,\text{m s}^{-2}$ . Find the exact time after $A$ passes $O$ when $B$ first catches $A$ , the distance from $O$ where this occurs, and the speed of $B$ relative to $A$ at that instant. Give all numerical answers exactly.
(6)
(Total for Question 4 is 6 marks)
5.
A train moves with constant acceleration. At $t=0$ the front of the train passes signal $A$ . At $t=4\,\text{s}$ the front passes signal $B$ , which is $60\,\text{m}$ beyond $A$ , with speed $18\,\text{m s}^{-1}$ . The rear of the train passes signal $A$ at $t=2\,\text{s}$ . Find the train's speed when its front passes $A$ , its acceleration, the exact length of the train and its speed when its rear passes $A$ . Give all numerical answers exactly.
(6)
(Total for Question 5 is 6 marks)

M7.4 · Use calculus in kinematics for motion in a straight line: v = dr/dt, a = dv/dt = d²r/dt², r = ∫v dt, v = ∫a dt; extend to 2 dimensions using vectors.

Explanation

Velocity is $\mathbf{v}=\dfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}$ and acceleration is $\mathbf{a}=\dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=\dfrac{\mathrm{d}^2\mathbf{r}}{\mathrm{d}t^2}$ , applied component by component for vectors.
Integrate acceleration to find velocity and integrate velocity to find position, using each given initial condition to determine the vector or scalar constant of integration.
To find distance rather than displacement, solve $v=0$ for direction changes and add the absolute changes in position across the resulting time intervals.
A common error is to omit constants of integration or to integrate speed instead of signed velocity without checking where the direction changes.

Worked example

A particle's velocity at time $t$ is $v=3t^2-12t+9\,\text{m s}^{-1}$ . It starts from position $2\,\text{m}$ . Determine its position function and the total distance it covers during $0\leq t\leq4$ .

1.Integrating gives $r=t^3-6t^2+9t+C$ ; $r(0)=2$ gives $C=2$ .
2.Since $v=3(t-1)(t-3)$ , direction changes occur at $t=1$ and $t=3$ .
3.The positions are $r(0)=2$ , $r(1)=6$ , $r(3)=2$ and $r(4)=6$ , so distance is $|6-2|+|2-6|+|6-2|=12\,\text{m}$ .

Answer: $r=t^3-6t^2+9t+2$ ; Total distance $=12\,\text{m}$

Common mistakes

Don't differentiate position once and label the result acceleration.
Don't integrate velocity to find displacement but equate net displacement with total distance despite direction changes.

Exam tip

For total distance, find every zero of velocity in the interval and add the magnitudes of the separate displacements.

Tier 1 · Easy

1.
A particle has position $r=t^3-4t^2+5t$ metres at time $t$ seconds. Find its velocity and acceleration at $t=2$ .
(3)
(Total for Question 1 is 3 marks)
2.
A particle has velocity $v=4t-3\,\text{m s}^{-1}$ . Find its displacement from $t=1$ to $t=3$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A particle has acceleration $a=6t-4\,\text{m s}^{-2}$ . Initially it is at position $-2\,\text{m}$ and moves at $3\,\text{m s}^{-1}$ . Find its velocity and position as functions of $t$ , and find both values when $t=2$ .
(5)
(Total for Question 1 is 5 marks)
2.
A particle moves on a straight line with acceleration $a=4-6t\,\text{m s}^{-2}$ . Given that $v=5\,\text{m s}^{-1}$ when $t=1$ and that its position is $2\,\text{m}$ when $t=0$ , find expressions for its velocity and position. Hence find its displacement from $t=0$ to $t=3$ .
(5)
(Total for Question 2 is 5 marks)
3.
A particle moves on a straight line with acceleration $a=6t-8\,\text{m s}^{-2}$ . Its velocity is $7\,\text{m s}^{-1}$ when $t=2$ . Find its velocity as a function of $t$ and find the exact minimum value of its velocity for $t\geq0$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A particle has acceleration $\mathbf{a}=(6t\mathbf{i}-4\mathbf{j})\,\text{m s}^{-2}$ . Initially its velocity is $(2\mathbf{i}+7\mathbf{j})\,\text{m s}^{-1}$ and its position vector is $(-\mathbf{i}+3\mathbf{j})\,\text{m}$ . Find its velocity and position vectors at time $t$ . Hence find its position and speed when its velocity is parallel to $\mathbf{i}$ .
(8)
(Total for Question 1 is 8 marks)
2.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. During the interval $0<t<1$ , the position of a moving particle is $\mathbf{r}=(t^2+2t)\mathbf{i}+(t^3-3t)\mathbf{j}$ metres. Find all the times in this interval when its acceleration is perpendicular to its velocity. Find the speed at each of these times, giving exact answers.
(6)
(Total for Question 2 is 6 marks)
3.
A particle moves on a straight line with acceleration $a=6t-6\,\text{m s}^{-2}$ . It starts at position $0\,\text{m}$ and returns to this position when $t=3\,\text{s}$ . Find its initial velocity. Hence find the time at which it is furthest from its initial position during $0\leq t\leq3$ , and find this greatest distance.
(6)
(Total for Question 3 is 6 marks)
4.
A particle moves on a straight line with acceleration $a=6t-4\,\text{m s}^{-2}$ . It is instantaneously at rest when $t=1\,\text{s}$ and its position is $5\,\text{m}$ when $t=2\,\text{s}$ . Find its velocity and position as functions of $t$ . Hence find all times in $0\leq t\leq2$ when the particle is at rest and find the exact total distance travelled during this interval. Give all numerical answers exactly.
(7)
(Total for Question 4 is 7 marks)
5.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are fixed perpendicular unit vectors. A particle has velocity $\mathbf{v}=(2t+1)\mathbf{i}+4\mathbf{j}\,\text{m s}^{-1}$ for $t\geq0$ and starts at the origin. Find its position vector at time $t$ and eliminate $t$ to obtain a Cartesian equation for its path, using coordinates $(x,y)$ . Find the first point where the path crosses the line $x=6$ , and find the exact speed there. Give all numerical answers exactly.
(6)
(Total for Question 5 is 6 marks)

M7.5 · Model motion under gravity in a vertical plane using vectors; projectiles.

Explanation

In the standard projectile model, the only acceleration is gravity vertically downwards, so horizontal velocity is constant while vertical velocity changes uniformly.
Resolve the initial velocity into horizontal and vertical components, then apply constant-acceleration equations separately with a clearly stated positive direction.
With launch speed $u$ at angle $\theta$ from level ground, eliminating $t$ from $x=u\cos\theta\,t$ and $y=u\sin\theta\,t-\dfrac12gt^2$ gives the equation of the path.
A common error is to use $v=0$ for the whole velocity at greatest height; only the vertical component is zero there, unless the projectile was launched vertically.

A projectile is resolved into horizontal

u\cos\theta

and vertical

u\sin\theta

components.

Worked example

A projectile is launched from level ground at $20\,\text{m s}^{-1}$ at $30^\circ$ above the horizontal. It lands at the same level. Using $g=9.8\,\text{m s}^{-2}$ , find its time of flight, horizontal range and greatest height.

1.The components are $u_x=20\cos30^\circ=10\sqrt3$ and $u_y=20\sin30^\circ=10$ .
2.From vertical displacement $0=10t-4.9t^2$ , the non-zero time is $t=10/4.9=2.0408\,\text{s}$ .
3.The range is $10\sqrt3(2.0408)=35.3\,\text{m}$ .
4.At the top, $0=10^2-2(9.8)h$ , so $h=100/19.6=5.10\,\text{m}$ .

Answer: Time of flight $=2.04\,\text{s}$ ; Range $=35.3\,\text{m}$ ; Greatest height $=5.10\,\text{m}$

Common mistakes

Don't use a same-level horizontal-range formula for a projectile that lands at a different vertical level.
Don't use the full launch speed in both directions instead of resolving it into horizontal and vertical components.

Exam tip

Resolve the initial velocity first, use zero vertical displacement for same-level flight, and keep horizontal acceleration zero.

Tier 1 · Easy

1.
A particle is projected vertically upwards at $14.7\,\text{m s}^{-1}$ . Using $g=9.8\,\text{m s}^{-2}$ , find the time taken to reach its greatest height and the height gained.
(4)
(Total for Question 1 is 4 marks)
2.
A particle is projected at $17\,\text{m s}^{-1}$ at an angle $\theta$ above the horizontal, where $\sin\theta=\dfrac{8}{17}$ . Find the horizontal and vertical components of its initial velocity.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A ball is projected horizontally at $12\,\text{m s}^{-1}$ from a point $25\,\text{m}$ above level ground. Using $g=9.8\,\text{m s}^{-2}$ and ignoring air resistance, find the time taken to reach the ground and the horizontal distance travelled, each to 3 significant figures.
(4)
(Total for Question 1 is 4 marks)
2.
A particle is projected from a point in a vertical plane. After $1.5\,\text{s}$ its velocity is $(14\mathbf{i}-2.7\mathbf{j})\,\text{m s}^{-1}$ , where $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertically upward unit vectors respectively. Assume that air resistance is negligible and take the downward acceleration as $9.8\,\text{m s}^{-2}$ . Find its initial speed and its angle of projection above the horizontal. Give both answers to 3 significant figures.
(4)
(Total for Question 2 is 4 marks)
3.
A particle is projected with velocity $(15\mathbf{i}+12\mathbf{j})\,\text{m s}^{-1}$ , where $\mathbf{i}$ is horizontal and $\mathbf{j}$ is vertically upwards. A vertical wall is $30\,\text{m}$ horizontally from the point of projection. Model the particle as moving freely under gravity and ignore air resistance. Take $g=9.8\,\text{m s}^{-2}$ . Find the height of the particle above the point of projection when it reaches the wall and find its velocity at that instant.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
From a point $5\,\text{m}$ above horizontal ground, a projectile is launched with velocity $(12\mathbf{i}+5\mathbf{j})\,\text{m s}^{-1}$ . Take $\mathbf{j}$ vertically upwards and use $g=10\,\text{m s}^{-2}$ . Find the Cartesian equation of its path, the horizontal distance to its first impact with the ground, and its speed on impact.
(8)
(Total for Question 1 is 8 marks)
2.
A particle is projected from horizontal ground with velocity $(18\mathbf{i}+24\mathbf{j})\,\text{m s}^{-1}$ , where $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertically upward unit vectors respectively. Assume that air resistance is negligible and use a constant downward acceleration of $9.8\,\text{m s}^{-2}$ . The particle passes twice through the horizontal level $20\,\text{m}$ above the point of projection. Find the exact horizontal distance between these two positions and the exact speed at the later position.
(6)
(Total for Question 2 is 6 marks)
3.
A particle is projected from horizontal ground and passes through the point $13\,\text{m}$ horizontally and $16.1\,\text{m}$ vertically above the point of projection after $1\,\text{s}$ . Model the particle as moving freely under gravity and ignore air resistance. Take $g=9.8\,\text{m s}^{-2}$ . Find its initial velocity. Hence find its greatest height and its horizontal distance from the point of projection when it returns to the ground, giving the distance to 3 significant figures. Use unrounded values in your working.
(6)
(Total for Question 3 is 6 marks)
4.
A particle is projected from a point with speed $14\,\text{m s}^{-1}$ at an angle $\theta$ above the horizontal, where $0<\theta<90^\circ$ . It passes through a point $20\,\text{m}$ horizontally from and $5\,\text{m}$ below the point of projection. Model the particle as moving freely under gravity, ignore air resistance and take $g=9.8\,\text{m s}^{-2}$ . Show that $2\tan^2\theta-4\tan\theta+1=0$ . Hence find all possible exact values of $\theta$ in the stated interval. Give all numerical answers exactly.
(6)
(Total for Question 4 is 6 marks)
5.
The vectors $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertically upward unit vectors respectively. A particle is projected from the origin with velocity $(8\mathbf{i}+15\mathbf{j})\,\text{m s}^{-1}$ . It first strikes a plane whose cross-section has equation $y=x/2$ . Model the particle as moving freely under gravity, ignore air resistance and take $g=9.8\,\text{m s}^{-2}$ . Find the exact time of impact, the distance measured along the plane from the origin to the impact point, and the exact speed of impact. Give all numerical answers exactly.
(6)
(Total for Question 5 is 6 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

M7.1 · Understand and use the language of kinematics: position; displacement; distance travelled; velocity; speed; acceleration.

Tier 1 · Easy

Mark scheme for M7.1 Tier 1 · Easy
Question	Scheme	Marks
1	Displacement $=4\,\text{m}$ Distance $=12\,\text{m}$	3
(3 marks)		3
Notes
The displacement is final position minus initial position: $1-(-3)=4\,\text{m}$ . The two path lengths are $5-(-3)=8\,\text{m}$ and $5-1=4\,\text{m}$ , so the distance is $8+4=12\,\text{m}$ .
2	Speed $=3.5\,\text{m s}^{-1}$ The lift is moving downwards	2
(2 marks)		2
Notes
Speed is the magnitude of velocity, so it is $\|-3.5\|=3.5\,\text{m s}^{-1}$ . The negative sign means the lift moves opposite to the positive direction, so it is moving downwards.

Tier 2 · Standard

Mark scheme for M7.1 Tier 2 · Standard
Question	Scheme	Marks
1	Speed $7\,\text{m s}^{-1}$ The speed is decreasing because velocity and acceleration have opposite signs (equivalently, the acceleration acts opposite to the motion).	3
(3 marks)		3
Notes
Speed is the magnitude of velocity, so it is $\|-7\|=7\,\text{m s}^{-1}$ . The velocity is negative while the acceleration is positive, so the acceleration acts opposite to the direction of motion. The magnitude of the velocity is therefore decreasing at that instant.
2	Acceleration $=2\,\text{m s}^{-2}$ Its speed is initially decreasing It changes direction	4
(4 marks)		4
Notes
The acceleration is $[2-(-6)]/4=2\,\text{m s}^{-2}$ . Initially the velocity and acceleration have opposite signs, so the speed decreases. Since the velocity changes continuously from negative to positive, it passes through zero and the particle changes direction.
3	Speed at $t=0$ is $6\,\text{m s}^{-1}$ Average velocity $=-2\,\text{m s}^{-1}$ The distance travelled is not known, so the average speed cannot be determined	4
(4 marks)		4
Notes
Speed is the magnitude of velocity, so the initial speed is $\|-6\|=6\,\text{m s}^{-1}$ . Average velocity is displacement divided by elapsed time, giving $-4/2=-2\,\text{m s}^{-1}$ . The displacement does not reveal whether the particle reversed direction, so it does not determine the total distance and hence cannot determine average speed.

Tier 3 · Hard

Mark scheme for M7.1 Tier 3 · Hard
Question	Scheme	Marks
1	Distance $=200\,\text{m}$ Displacement $=100\,\text{m}$ east Average speed $=\dfrac{20}{3}\,\text{m s}^{-1}$ Average velocity $=\dfrac{10}{3}\,\text{m s}^{-1}$ east	6
(6 marks)		6
Notes
Distance adds all path lengths: $120+50+30=200\,\text{m}$ . Taking east as positive, displacement is $120-50+30=100\,\text{m}$ . The total time is $30\,\text{s}$ , so average speed is $200/30=20/3\,\text{m s}^{-1}$ and average velocity is $100/30=10/3\,\text{m s}^{-1}$ east.
2	Final position $=22\,\text{m}$ Distance $=48\,\text{m}$ Average speed $=\dfrac{48}{13}\,\text{m s}^{-1}$ Average velocity $=\dfrac{12}{13}\,\text{m s}^{-1}$ in the positive direction	6
(6 marks)		6
Notes
The signed displacements are $-3(4)=-12$ , $5(6)=30$ and $-2(3)=-6$ metres. Their sum is $12\,\text{m}$ , so the final position is $10+12=22\,\text{m}$ . Distance is $12+30+6=48\,\text{m}$ and total time is $13\,\text{s}$ . Hence average speed is $48/13\,\text{m s}^{-1}$ and average velocity is $12/13\,\text{m s}^{-1}$ .
3	Distance in the positive direction $=20\,\text{m}$ Distance in the negative direction $=8\,\text{m}$ The particle changes direction at position $15\,\text{m}$ Average velocity $=1.5\,\text{m s}^{-1}$ in the positive direction	6
(6 marks)		6
Notes
The total distance is $3.5(8)=28\,\text{m}$ and the displacement is $7-(-5)=12\,\text{m}$ . If $p$ and $n$ are the distances in the positive and negative directions, then $p+n=28$ and $p-n=12$ . Hence $p=20$ and $n=8$ . The single change of direction occurs after the positive motion, at position $-5+20=15\,\text{m}$ . The average velocity is $12/8=1.5\,\text{m s}^{-1}$ in the positive direction.
4	Displacement $=(18\mathbf{i}+24\mathbf{j})\,\text{m}$ Displacement magnitude $=30\,\text{m}$ Average speed $=3\,\text{m s}^{-1}$ Average velocity $=(\dfrac97\mathbf{i}+\dfrac{12}{7}\mathbf{j})\,\text{m s}^{-1}$ Its magnitude is $15/7\,\text{m s}^{-1}$ , less than the average speed because distance exceeds displacement magnitude	6
(6 marks)		6
Notes
Final position minus initial position gives $(16-(-2))\mathbf{i}+(25-1)\mathbf{j}=18\mathbf{i}+24\mathbf{j}$ metres, with magnitude $\sqrt{18^2+24^2}=30\,\text{m}$ . Average speed is distance divided by time, $42/14=3\,\text{m s}^{-1}$ . Average velocity is displacement divided by time, giving $(9/7)\mathbf{i}+(12/7)\mathbf{j}\,\text{m s}^{-1}$ . Its magnitude is $30/14=15/7\,\text{m s}^{-1}$ ; the winding path length $42\,\text{m}$ exceeds the $30\,\text{m}$ endpoint separation.
5	Average velocity $=(4\mathbf{i}-2\mathbf{j})\,\text{m s}^{-1}$ Average acceleration $=(\dfrac43\mathbf{i}-2\mathbf{j})\,\text{m s}^{-2}$ Initial speed $=5\,\text{m s}^{-1}$ Final speed $=\sqrt{89}\,\text{m s}^{-1}$ The path between the endpoints is not specified, so its length is not determined	6
(6 marks)		6
Notes
The displacement is $(20-(-4))\mathbf{i}+(-9-3)\mathbf{j}=24\mathbf{i}-12\mathbf{j}$ metres, so average velocity is $(24\mathbf{i}-12\mathbf{j})/6=4\mathbf{i}-2\mathbf{j}\,\text{m s}^{-1}$ . The velocity change is $8\mathbf{i}-12\mathbf{j}$ , giving average acceleration $(4/3)\mathbf{i}-2\mathbf{j}\,\text{m s}^{-2}$ . The endpoint speeds are $\sqrt{(-3)^2+4^2}=5$ and $\sqrt{5^2+(-8)^2}=\sqrt{89}\,\text{m s}^{-1}$ . Endpoint positions and velocities do not specify the intervening path, so they cannot fix its total length.

M7.2 · Understand, use and interpret graphs in kinematics for motion in a straight line: displacement against time and interpretation of gradient; velocity against time and interpretation of gradient and area under the graph.

Tier 1 · Easy

Mark scheme for M7.2 Tier 1 · Easy
Question	Scheme	Marks
1	$3\,\text{m s}^{-1}$	2
(2 marks)		2
Notes
Velocity is the gradient of the displacement-time graph: $v=(23-5)/(8-2)=18/6=3\,\text{m s}^{-1}$ .
2	$14\,\text{m}$	2
(2 marks)		2
Notes
Displacement is the area under the velocity-time graph. The area is a triangle, so it is $\tfrac12(4)(7)=14\,\text{m}$ .

Tier 2 · Standard

Mark scheme for M7.2 Tier 2 · Standard
Question	Scheme	Marks
1	Velocities $5\,\text{m s}^{-1}$ and $-2\,\text{m s}^{-1}$ Total distance $32\,\text{m}$	4
(4 marks)		4
Notes
Velocity is the gradient of a displacement-time graph. The first gradient is $(20-0)/(4-0)=5\,\text{m s}^{-1}$ ; the second is $(8-20)/(10-4)=-12/6=-2\,\text{m s}^{-1}$ . The particle travels $20\,\text{m}$ out and then $12\,\text{m}$ back, so total distance is $32\,\text{m}$ .
2	$V=5\,\text{m s}^{-1}$ First acceleration $=0.5\,\text{m s}^{-2}$ Second acceleration $=-\dfrac23\,\text{m s}^{-2}$	5
(5 marks)		5
Notes
Using trapezium areas, $\tfrac12(3+V)(4)+\tfrac12(V+1)(6)=34$ . Thus $2(3+V)+3(V+1)=34$ , so $5V+9=34$ and $V=5$ . The gradients give accelerations $(5-3)/4=0.5\,\text{m s}^{-2}$ and $(1-5)/6=-2/3\,\text{m s}^{-2}$ .
3	Velocity at $t=4$ is $-3\,\text{m s}^{-1}$ Velocity at $t=9$ is zero, so the particle is instantaneously at rest	4
(4 marks)		4
Notes
Velocity is the gradient of the displacement-time graph. The gradient of the tangent at $t=4$ is $(-4-11)/(7-2)=-15/5=-3\,\text{m s}^{-1}$ . A horizontal tangent has gradient zero, so at $t=9$ the velocity is zero and the particle is instantaneously at rest.

Tier 3 · Hard

Mark scheme for M7.2 Tier 3 · Hard
Question	Scheme	Marks
1	Final acceleration $=-3\,\text{m s}^{-2}$ Displacement $=57\,\text{m}$ Distance $=\dfrac{175}{3}\,\text{m}$	7
(7 marks)		7
Notes
The final gradient is $(-2-10)/(9-5)=-3\,\text{m s}^{-2}$ . The signed areas are $\tfrac12(4+10)(3)=21$ , $10(2)=20$ and $\tfrac12(10-2)(4)=16$ , so displacement is $57\,\text{m}$ . In the final stage velocity reaches zero after $10/3\,\text{s}$ ; its positive and negative area magnitudes are $50/3$ and $2/3$ . Hence distance is $21+20+50/3+2/3=175/3\,\text{m}$ .
2	$T=10\,\text{s}$ Final velocity $=-4\,\text{m s}^{-1}$ The particle changes direction at $t=\dfrac{20}{3}\,\text{s}$	6
(6 marks)		6
Notes
Let the positive and negative area magnitudes be $A$ and $B$ . Then $A-B=20$ and $A+B=100/3$ , giving $A=80/3$ and $B=20/3$ . The graph consists of similar triangles, so $B/A=(\|v_T\|/8)^2=1/4$ , hence $v_T=-4\,\text{m s}^{-1}$ . The whole trapezium has signed area $\tfrac12(8-4)T=2T=20$ , so $T=10$ . The velocity reaches zero after the fraction $8/(8+4)=2/3$ of the interval, at $t=20/3\,\text{s}$ .
3	Acceleration $=-0.5\,\text{m s}^{-2}$ Displacement $=4-\dfrac{9\pi}{2}\,\text{m}$ Distance $=4+\dfrac{9\pi}{2}\,\text{m}$	6
(6 marks)		6
Notes
The gradient of the straight segment is $(0-2)/(4-0)=-0.5\,\text{m s}^{-2}$ . Its area above the axis is $\tfrac12(4)(2)=4\,\text{m}$ . The semicircle has radius $3$ in the graph coordinates, so the magnitude of its area is $\tfrac12\pi(3^2)=9\pi/2\,\text{m}$ . This area is below the axis, giving displacement $4-9\pi/2\,\text{m}$ . Distance uses both area magnitudes, giving $4+9\pi/2\,\text{m}$ .
4	$k=14$ Velocities $4\,\text{m s}^{-1}$ and $-\dfrac{18}{5}\,\text{m s}^{-1}$ Average speed $=\dfrac{15}{4}\,\text{m s}^{-1}$ Average velocity $=-\dfrac34\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
Because $k>2$ and the final displacement is $-4$ , the two distances are $k-2$ and $k+4$ . Thus $(k-2)+(k+4)=30$ , so $2k+2=30$ and $k=14$ . The segment gradients are $(14-2)/3=4\,\text{m s}^{-1}$ and $(-4-14)/(8-3)=-18/5\,\text{m s}^{-1}$ . Average speed is $30/8=15/4\,\text{m s}^{-1}$ , while average velocity is the net displacement $-4-2=-6$ divided by $8$ , giving $-3/4\,\text{m s}^{-1}$ .
5	Meeting times $t=2\,\text{s}$ and $t=6\,\text{s}$ Common positions $12\,\text{m}$ and $36\,\text{m}$ Immediately after the first meeting $A$ is faster; immediately after the second meeting $B$ is faster	7
(7 marks)		7
Notes
The area under $A$ 's graph gives $x_A=6t$ . For $B$ , the area under its line from velocity $2$ to $2+t$ is $\tfrac12[2+(2+t)]t=2t+t^2/2$ , so $x_B=6+2t+t^2/2$ . Equating positions gives $6t=6+2t+t^2/2$ , hence $t^2-8t+12=0$ and $t=2$ or $6$ . The common positions are $6t$ , namely $12$ and $36\,\text{m}$ . Particle $B$ has velocity $2+t$ : at $t=2$ this is $4<6$ , while at $t=6$ it is $8>6$ .

M7.3 · Understand, use and derive the formulae for constant acceleration for motion in a straight line; extend to 2 dimensions using vectors.

Tier 1 · Easy

Mark scheme for M7.3 Tier 1 · Easy
Question	Scheme	Marks
1	$2\,\text{m s}^{-2}$	2
(2 marks)		2
Notes
From $v=u+at$ , $11=5+3a$ , so $a=(11-5)/3=2\,\text{m s}^{-2}$ .
2	$60\,\text{m}$	2
(2 marks)		2
Notes
$s=ut+\tfrac12at^2=9(5)+\tfrac12(1.2)(5^2)=45+15=60\,\text{m}$ .

Tier 2 · Standard

Mark scheme for M7.3 Tier 2 · Standard
Question	Scheme	Marks
1	Velocity $(13\mathbf{i}-\mathbf{j})\,\text{m s}^{-1}$ Displacement $(40\mathbf{i}+7.5\mathbf{j})\,\text{m}$	4
(4 marks)		4
Notes
Use the constant-acceleration formulae componentwise. The velocity is $\mathbf{v}=\mathbf{u}+t\mathbf{a}=(3,4)+5(2,-1)=(13,-1)$ . The displacement is $\mathbf{s}=t\mathbf{u}+\tfrac12t^2\mathbf{a}=5(3,4)+12.5(2,-1)=(40,7.5)$ .
2	Initial velocity $=(2\mathbf{i}-5\mathbf{j})\,\text{m s}^{-1}$ Velocity at $t=4$ is $(10\mathbf{i}+7\mathbf{j})\,\text{m s}^{-1}$	4
(4 marks)		4
Notes
Using $\mathbf{s}=\mathbf{u}t+\tfrac12\mathbf{a}t^2$ , $(24,4)=4\mathbf{u}+8(2,3)=4\mathbf{u}+(16,24)$ . Hence $4\mathbf{u}=(8,-20)$ and $\mathbf{u}=(2,-5)$ . Then $\mathbf{v}=\mathbf{u}+4\mathbf{a}=(2,-5)+(8,12)=(10,7)$ .
3	$v^2=u^2+2as$ Stopping distance $=32.5\,\text{m}$	5
(5 marks)		5
Notes
From $v=u+at$ , $t=(v-u)/a$ . Substitution in $s=\tfrac12(u+v)t$ gives $s=(u+v)(v-u)/(2a)=(v^2-u^2)/(2a)$ , so $v^2=u^2+2as$ . With $v=0$ , $u=13$ and $a=-2.6$ , $0=13^2+2(-2.6)s$ , hence $s=169/5.2=32.5\,\text{m}$ .

Tier 3 · Hard

Mark scheme for M7.3 Tier 3 · Hard
Question	Scheme	Marks
1	$t=4\,\text{s}$ Displacement $=(28\mathbf{i}+16\mathbf{j})\,\text{m}$ Speed $=11\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
$\mathbf{v}=\mathbf{u}+\mathbf{a}t=(3+2t)\mathbf{i}+(8-2t)\mathbf{j}$ . It is parallel to $\mathbf{i}$ when $8-2t=0$ , giving $t=4$ . Then $\mathbf{r}=\mathbf{u}t+\tfrac12\mathbf{a}t^2=(12\mathbf{i}+32\mathbf{j})+(16\mathbf{i}-16\mathbf{j})=28\mathbf{i}+16\mathbf{j}$ . The velocity is $11\mathbf{i}\,\text{m s}^{-1}$ , so the speed is $11\,\text{m s}^{-1}$ .
2	Velocity $=(3\mathbf{i}-3\mathbf{j})\,\text{m s}^{-1}$ Position $=(\dfrac{53}{2}\mathbf{i}-14\mathbf{j})\,\text{m}$	6
(6 marks)		6
Notes
After $3\,\text{s}$ , $\mathbf{v}_1=(4,1)+3(1,-2)=(7,-5)$ and $\mathbf{r}_1=3(4,1)+\tfrac12(3^2)(1,-2)=(33/2,-6)$ . During the next $2\,\text{s}$ , $\mathbf{v}_2=(7,-5)+2(-2,1)=(3,-3)$ . The second displacement is $2(7,-5)+\tfrac12(2^2)(-2,1)=(10,-8)$ , so $\mathbf{r}_2=(33/2,-6)+(10,-8)=(53/2,-14)$ .
3	Initial velocity $=(3\mathbf{i}-\mathbf{j})\,\text{m s}^{-1}$ Acceleration $=(2\mathbf{i}+4\mathbf{j})\,\text{m s}^{-2}$ Velocity after $5\,\text{s}$ is $(13\mathbf{i}+19\mathbf{j})\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
Write the initial velocity as $\mathbf{u}$ and acceleration as $\mathbf{a}$ . From $\mathbf{r}=\mathbf{u}t+\tfrac12\mathbf{a}t^2$ , $2\mathbf{u}+2\mathbf{a}=10\mathbf{i}+6\mathbf{j}$ and $5\mathbf{u}+\tfrac{25}{2}\mathbf{a}=40\mathbf{i}+45\mathbf{j}$ . Thus $\mathbf{u}+\mathbf{a}=5\mathbf{i}+3\mathbf{j}$ . Substituting $\mathbf{u}=5\mathbf{i}+3\mathbf{j}-\mathbf{a}$ into the second equation gives $\tfrac{15}{2}\mathbf{a}=15\mathbf{i}+30\mathbf{j}$ , so $\mathbf{a}=2\mathbf{i}+4\mathbf{j}$ and $\mathbf{u}=3\mathbf{i}-\mathbf{j}$ . Hence $\mathbf{v}=\mathbf{u}+5\mathbf{a}=13\mathbf{i}+19\mathbf{j}$ .
4	Catch time $=\dfrac{9+\sqrt{65}}{2}\,\text{s}$ Distance from $O=5(9+\sqrt{65})\,\text{m}$ Relative speed $=2\sqrt{65}\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
At global time $t\geq2$ , $A$ is $10t$ metres from $O$ , while $B$ has moved $\tfrac12(4)(t-2)^2=2(t-2)^2$ metres. Equating gives $2(t-2)^2=10t$ , or $t^2-9t+4=0$ . The root exceeding $2$ is $(9+\sqrt{65})/2$ . The meeting distance is $10t=5(9+\sqrt{65})\,\text{m}$ . Particle $B$ then has speed $4(t-2)=2(5+\sqrt{65})=10+2\sqrt{65}$ , so its speed relative to $A$ is $2\sqrt{65}\,\text{m s}^{-1}$ .
5	Initial speed $=12\,\text{m s}^{-1}$ Acceleration $=1.5\,\text{m s}^{-2}$ Train length $=27\,\text{m}$ Speed when the rear passes $A=15\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
Let the speed at $t=0$ be $u$ and the acceleration be $a$ . Then $18=u+4a$ and $60=4u+8a$ . Substituting $u=18-4a$ into the second equation gives $60=72-8a$ , so $a=1.5$ and $u=12$ . Between $t=0$ and $t=2$ , the front travels $12(2)+\tfrac12(1.5)(2^2)=27\,\text{m}$ ; this is the train's length because the rear then reaches $A$ . Its speed at that time is $12+1.5(2)=15\,\text{m s}^{-1}$ .

M7.4 · Use calculus in kinematics for motion in a straight line: v = dr/dt, a = dv/dt = d²r/dt², r = ∫v dt, v = ∫a dt; extend to 2 dimensions using vectors.

Tier 1 · Easy

Mark scheme for M7.4 Tier 1 · Easy
Question	Scheme	Marks
1	Velocity $=1\,\text{m s}^{-1}$ Acceleration $=4\,\text{m s}^{-2}$	3
(3 marks)		3
Notes
$v=\dfrac{\mathrm{d}r}{\mathrm{d}t}=3t^2-8t+5$ , so $v(2)=12-16+5=1\,\text{m s}^{-1}$ . Then $a=\dfrac{\mathrm{d}v}{\mathrm{d}t}=6t-8$ , giving $a(2)=4\,\text{m s}^{-2}$ .
2	$10\,\text{m}$	2
(2 marks)		2
Notes
The displacement is $\int_1^3(4t-3)\,\mathrm{d}t=[2t^2-3t]_1^3=(18-9)-(2-3)=10\,\text{m}$ .

Tier 2 · Standard

Mark scheme for M7.4 Tier 2 · Standard
Question	Scheme	Marks
1	$v=3t^2-4t+3$ $r=t^3-2t^2+3t-2$ $v(2)=7\,\text{m s}^{-1}$ and $r(2)=4\,\text{m}$	5
(5 marks)		5
Notes
Integrate acceleration: $v=3t^2-4t+C$ . Since $v(0)=3$ , $C=3$ . Integrate again: $r=t^3-2t^2+3t+D$ . Since $r(0)=-2$ , $D=-2$ . Substitution of $t=2$ gives $v(2)=12-8+3=7$ and $r(2)=8-8+6-2=4$ .
2	$v=4t-3t^2+4$ $r=2t^2-t^3+4t+2$ Displacement $=3\,\text{m}$	5
(5 marks)		5
Notes
Integrating gives $v=4t-3t^2+C$ . Since $v(1)=5$ , $1+C=5$ and $C=4$ . Integrating again gives $r=2t^2-t^3+4t+D$ . Since $r(0)=2$ , $D=2$ . Thus $r(3)=18-27+12+2=5$ , so the displacement is $r(3)-r(0)=5-2=3\,\text{m}$ .
3	$v=3t^2-8t+11\,\text{m s}^{-1}$ Minimum velocity $=\dfrac{17}{3}\,\text{m s}^{-1}$ , attained when $t=\dfrac43\,\text{s}$	4
(4 marks)		4
Notes
Integrating the acceleration gives $v=3t^2-8t+C$ . Since $v=7$ at $t=2$ , $7=12-16+C$ , so $C=11$ . The velocity is stationary when $dv/dt=a=0$ , giving $6t-8=0$ and $t=4/3$ . As the quadratic coefficient is positive this is a minimum, with $v=3(16/9)-8(4/3)+11=17/3\,\text{m s}^{-1}$ .

Tier 3 · Hard

Mark scheme for M7.4 Tier 3 · Hard
Question	Scheme	Marks
1	$\mathbf{v}=(3t^2+2)\mathbf{i}+(7-4t)\mathbf{j}$ $\mathbf{r}=(t^3+2t-1)\mathbf{i}+(3+7t-2t^2)\mathbf{j}$ At $t=\dfrac74$ , $\mathbf{r}=\dfrac{503}{64}\mathbf{i}+\dfrac{73}{8}\mathbf{j}$ metres Speed $=\dfrac{179}{16}\,\text{m s}^{-1}$	8
(8 marks)		8
Notes
Integrating $\mathbf{a}$ and using $\mathbf{v}(0)=2\mathbf{i}+7\mathbf{j}$ gives $\mathbf{v}=(3t^2+2)\mathbf{i}+(7-4t)\mathbf{j}$ . Integrating again and using $\mathbf{r}(0)=-\mathbf{i}+3\mathbf{j}$ gives $\mathbf{r}=(t^3+2t-1)\mathbf{i}+(3+7t-2t^2)\mathbf{j}$ . Parallel to $\mathbf{i}$ requires $7-4t=0$ , so $t=7/4$ . Substitution gives $\mathbf{r}=503\mathbf{i}/64+73\mathbf{j}/8$ . The remaining velocity component is $3(7/4)^2+2=179/16$ , so the speed is $179/16\,\text{m s}^{-1}$ .
2	$t=\dfrac13$ and $t=\dfrac23$ At $t=\dfrac13$ , speed $=\dfrac{8\sqrt2}{3}\,\text{m s}^{-1}$ At $t=\dfrac23$ , speed $=\dfrac{5\sqrt5}{3}\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
$\mathbf{v}=(2t+2)\mathbf{i}+(3t^2-3)\mathbf{j}$ and $\mathbf{a}=2\mathbf{i}+6t\mathbf{j}$ . Perpendicular vectors have zero scalar product, so $2(2t+2)+6t(3t^2-3)=18t^3-14t+4=2(3t-1)(3t-2)(t+1)=0$ . In $0<t<1$ , $t=1/3$ or $2/3$ . The velocities are $(8/3,-8/3)$ and $(10/3,-5/3)$ , whose magnitudes are $8\sqrt2/3$ and $5\sqrt5/3\,\text{m s}^{-1}$ respectively.
3	Initial velocity $=0\,\text{m s}^{-1}$ The particle is furthest from its initial position at $t=2\,\text{s}$ Greatest distance $=4\,\text{m}$	6
(6 marks)		6
Notes
Integrating gives $v=3t^2-6t+C$ and displacement $s=t^3-3t^2+Ct$ , since $s=0$ at $t=0$ . The return condition $s(3)=0$ gives $27-27+3C=0$ , so $C=0$ and the initial velocity is zero. Stationary positions occur when $v=3t(t-2)=0$ , so the candidates in the closed interval are $t=0,2,3$ . Their displacements are $0,-4,0$ metres respectively, so the particle is furthest from its initial position at $t=2$ and the greatest distance is $4\,\text{m}$ .
4	$v=3t^2-4t+1\,\text{m s}^{-1}$ $r=t^3-2t^2+t+3\,\text{m}$ Rest times $t=\dfrac13\,\text{s}$ and $t=1\,\text{s}$ Total distance $=\dfrac{62}{27}\,\text{m}$	7
(7 marks)		7
Notes
Integrating gives $v=3t^2-4t+C$ . Since $v(1)=0$ , $C=1$ . Integrating again gives $r=t^3-2t^2+t+D$ ; $r(2)=5$ gives $D=3$ . Now $v=(3t-1)(t-1)$ , so the rest times are $1/3$ and $1$ . The positions at $t=0,1/3,1,2$ are $3,85/27,3,5$ . Hence the distance is $\|85/27-3\|+\|3-85/27\|+\|5-3\|=4/27+4/27+2=62/27\,\text{m}$ .
5	$\mathbf{r}=(t^2+t)\mathbf{i}+4t\mathbf{j}\,\text{m}$ $x=\dfrac{y^2}{16}+\dfrac{y}{4}$ First crossing point $(6,8)$ Speed $=\sqrt{41}\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
Integrating the components and using $\mathbf r(0)=\mathbf0$ gives $x=t^2+t$ and $y=4t$ . Thus $t=y/4$ and $x=y^2/16+y/4$ . On $x=6$ , $t^2+t=6$ , so $(t-2)(t+3)=0$ ; the allowed time is $t=2$ , giving $y=8$ . The velocity then is $5\mathbf{i}+4\mathbf{j}\,\text{m s}^{-1}$ , whose magnitude is $\sqrt{25+16}=\sqrt{41}\,\text{m s}^{-1}$ .

M7.5 · Model motion under gravity in a vertical plane using vectors; projectiles.

Tier 1 · Easy

Mark scheme for M7.5 Tier 1 · Easy
Question	Scheme	Marks
1	Time $=1.5\,\text{s}$ Height gained $=11.025\,\text{m}$	4
(4 marks)		4
Notes
Taking upwards as positive, $v=u-gt$ . At greatest height $v=0$ , so $0=14.7-9.8t$ and $t=1.5\,\text{s}$ . Then $v^2=u^2+2as$ gives $0=14.7^2-2(9.8)s$ , so $s=11.025\,\text{m}$ .
2	Horizontal component $=15\,\text{m s}^{-1}$ Vertical component $=8\,\text{m s}^{-1}$ upwards	2
(2 marks)		2
Notes
Since $\theta$ is acute, $\cos\theta=15/17$ . Therefore the components are $17\cos\theta=15\,\text{m s}^{-1}$ horizontally and $17\sin\theta=8\,\text{m s}^{-1}$ vertically upwards.

Tier 2 · Standard

Mark scheme for M7.5 Tier 2 · Standard
Question	Scheme	Marks
1	Time $=2.26\,\text{s}$ Horizontal distance $=27.1\,\text{m}$	4
(4 marks)		4
Notes
Vertically the initial velocity is zero, so $25=\tfrac12(9.8)t^2$ . Hence $t=\sqrt{50/9.8}=2.2587\ldots\,\text{s}$ , which is $2.26\,\text{s}$ to 3 significant figures. Horizontal velocity remains $12\,\text{m s}^{-1}$ , so the horizontal distance is $12t=27.105\ldots\,\text{m}$ , or $27.1\,\text{m}$ to 3 significant figures. Keep the unrounded $t$ on your calculator for that multiplication rather than retyping a rounded value.
2	Initial speed $=18.4\,\text{m s}^{-1}$ Angle of projection $=40.6^\circ$	4
(4 marks)		4
Notes
Horizontal velocity is constant, so the initial horizontal component is $14$ . Vertically, $-2.7=u_y-9.8(1.5)$ , giving $u_y=12$ . Hence the initial speed is $\sqrt{14^2+12^2}=2\sqrt{85}=18.4\,\text{m s}^{-1}$ to 3 significant figures. The angle is $\tan^{-1}(12/14)=40.6^\circ$ to 3 significant figures.
3	Height $=4.4\,\text{m}$ Velocity $=(15\mathbf{i}-7.6\mathbf{j})\,\text{m s}^{-1}$	5
(5 marks)		5
Notes
Horizontal motion gives $30=15t$ , so the particle reaches the wall when $t=2\,\text{s}$ . Its vertical displacement is $12(2)-\tfrac12(9.8)(2^2)=4.4\,\text{m}$ . The horizontal velocity remains $15\,\text{m s}^{-1}$ and the vertical velocity is $12-9.8(2)=-7.6\,\text{m s}^{-1}$ , giving the stated velocity vector.

Tier 3 · Hard

Mark scheme for M7.5 Tier 3 · Hard
Question	Scheme	Marks
1	$y=5+\dfrac{5}{12}x-\dfrac{5}{144}x^2$ Horizontal distance $=6(1+\sqrt5)\,\text{m}$ Impact speed $=\sqrt{269}\,\text{m s}^{-1}$	8
(8 marks)		8
Notes
$x=12t$ and $y=5+5t-5t^2$ . Substituting $t=x/12$ gives $y=5+5x/12-5x^2/144$ . At impact, $5+5t-5t^2=0$ , so $t=(1+\sqrt5)/2$ and $x=12t=6(1+\sqrt5)\,\text{m}$ . The impact velocity is $12\mathbf{i}+(5-10t)\mathbf{j}=12\mathbf{i}-5\sqrt5\mathbf{j}$ , so its speed is $\sqrt{12^2+(5\sqrt5)^2}=\sqrt{269}\,\text{m s}^{-1}$ .
2	Horizontal distance $=\dfrac{360\sqrt{46}}{49}\,\text{m}$ Later speed $=2\sqrt{127}\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
The vertical displacement is $y=24t-4.9t^2$ . Setting $y=20$ gives $9.8t^2-48t+40=0$ , whose roots differ by $\sqrt{48^2-4(9.8)(40)}/9.8=2\sqrt{184}/9.8=20\sqrt{46}/49$ . Horizontal speed is constant at $18\,\text{m s}^{-1}$ , so the horizontal separation is $18(20\sqrt{46}/49)=360\sqrt{46}/49\,\text{m}$ . At the later crossing the vertical velocity is $-\sqrt{184}=-2\sqrt{46}\,\text{m s}^{-1}$ , so the speed is $\sqrt{18^2+(2\sqrt{46})^2}=\sqrt{508}=2\sqrt{127}\,\text{m s}^{-1}$ .
3	Initial velocity $=(13\mathbf{i}+21\mathbf{j})\,\text{m s}^{-1}$ Greatest height $=22.5\,\text{m}$ Horizontal distance $=55.7\,\text{m}$	6
(6 marks)		6
Notes
Horizontal motion gives $13=u_x(1)$ , so $u_x=13\,\text{m s}^{-1}$ . Vertically, $16.1=u_y(1)-\tfrac12(9.8)(1^2)$ , so $u_y=21\,\text{m s}^{-1}$ . At greatest height, $0=21^2-2(9.8)H$ , giving $H=22.5\,\text{m}$ . The non-zero solution of $0=21t-4.9t^2$ is $t=30/7\,\text{s}$ , so the horizontal distance is $13(30/7)=390/7=55.714\ldots\,\text{m}$ , which is $55.7\,\text{m}$ to 3 significant figures.
4	$2\tan^2\theta-4\tan\theta+1=0$ $\theta=\tan^{-1}(1-\dfrac{\sqrt2}{2})$ or $\theta=\tan^{-1}(1+\dfrac{\sqrt2}{2})$	6
(6 marks)		6
Notes
Write $x=14\cos\theta\,t$ and $y=14\sin\theta\,t-4.9t^2$ . Eliminating $t$ gives $y=x\tan\theta-[9.8x^2/(2(14)^2)]\sec^2\theta$ . At $x=20$ the coefficient is $10$ , so $-5=20T-10(1+T^2)$ where $T=\tan\theta$ . This simplifies to $2T^2-4T+1=0$ , giving $T=1\pm\sqrt2/2$ . Both values are positive, so both inverse-tangent values lie in the stated interval.
5	Impact time $=\dfrac{110}{49}\,\text{s}$ Distance along the plane $=\dfrac{440\sqrt5}{49}\,\text{m}$ Impact speed $=\sqrt{113}\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
$x=8t$ and $y=15t-4.9t^2$ . At the plane, $15t-4.9t^2=(8t)/2=4t$ . Besides $t=0$ , this gives $t=11/4.9=110/49\,\text{s}$ . The impact coordinates are $(880/49,440/49)$ , so the distance from the origin along the plane is $\sqrt{(880/49)^2+(440/49)^2}=440\sqrt5/49\,\text{m}$ . The impact velocity is $8\mathbf{i}+[15-9.8(110/49)]\mathbf{j}=8\mathbf{i}-7\mathbf{j}$ , with speed $\sqrt{113}\,\text{m s}^{-1}$ .