M7 Kinematics — 9MA0 revision notes

M7.1

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

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

M7.3

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

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

M7.5

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

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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