M7 Kinematics — printable coverage pack

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

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.

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 marks]
Answer
- Displacement $=4\,\text{m}$
- Distance $=12\,\text{m}$
Method: 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 $8+4=12\,\text{m}$ .

Tier 2 · Standard

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

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 marks]
Answer
- 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
Method: 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.

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.

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.

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 marks]
Answer
- $3\,\text{m s}^{-1}$
Method: Velocity is the gradient of the displacement-time graph: $v=(23-5)/(8-2)=18/6=3\,\text{m s}^{-1}$ .

Tier 2 · Standard

1. 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 $7\,\text{s}$ .[4 marks]
Answer
- Acceleration $=2\,\text{m s}^{-2}$
- Displacement $=47\,\text{m}$
Method: The first gradient is $(8-2)/3=2\,\text{m s}^{-2}$ . 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 $15+32=47\,\text{m}$ .

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 marks]
Answer
- Final acceleration $=-3\,\text{m s}^{-2}$
- Displacement $=57\,\text{m}$
- Distance $=\dfrac{175}{3}\,\text{m}$
Method: 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 $21+20+50/3+2/3=175/3\,\text{m}$ .

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

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.

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 marks]
Answer
- $2\,\text{m s}^{-2}$
Method: From $v=u+at$ , $11=5+3a$ , so $a=(11-5)/3=2\,\text{m s}^{-2}$ .

Tier 2 · Standard

1. 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 $6\,\text{s}$ .[4 marks]
Answer
- Speed $=13\,\text{m s}^{-1}$
- Distance $=51\,\text{m}$
Method: $v=u+at=4+1.5(6)=13\,\text{m s}^{-1}$ . Also $s=ut+\tfrac12at^2=4(6)+\tfrac12(1.5)(6^2)=24+27=51\,\text{m}$ .

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 marks]
Answer
- $t=4\,\text{s}$
- Displacement $=(28\mathbf{i}+16\mathbf{j})\,\text{m}$
- Speed $=11\,\text{m s}^{-1}$
Method: $\mathbf{v}=\mathbf{u}+\mathbf{a}t=(3+2t)\mathbf{i}+(8-2t)\mathbf{j}$ . 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 $11\,\text{m s}^{-1}$ .

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.

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.

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 marks]
Answer
- Velocity $=1\,\text{m s}^{-1}$
- Acceleration $=4\,\text{m s}^{-2}$
Method: $v=\dfrac{\mathrm{d}r}{\mathrm{d}t}=3t^2-8t+5$ , so $v(2)=12-16+5=1\,\text{m s}^{-1}$ . Then $a=\dfrac{\mathrm{d}v}{\mathrm{d}t}=6t-8$ , giving $a(2)=4\,\text{m s}^{-2}$ .

Tier 2 · Standard

1. 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 $0\leq t\leq4$ .[6 marks]
Answer
- $r=t^3-6t^2+9t+2$
- Total distance $=12\,\text{m}$
Method: Integrating gives $r=t^3-6t^2+9t+C$ ; $r(0)=2$ gives $C=2$ . Since $v=3(t-1)(t-3)$ , direction changes occur at $t=1$ and $t=3$ . The positions are $r(0)=2$ , $r(1)=6$ , $r(3)=2$ and $r(4)=6$ , so distance is $|6-2|+|2-6|+|6-2|=12\,\text{m}$ .

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 marks]
Answer
- $\mathbf{v}=(3t^2+2)\mathbf{i}+(7-4t)\mathbf{j}$
- $\mathbf{r}=(t^3+2t-1)\mathbf{i}+(3+7t-2t^2)\mathbf{j}$
- At $t=\dfrac74$ , $\mathbf{r}=\dfrac{503}{64}\mathbf{i}+\dfrac{73}{8}\mathbf{j}$ metres
- Speed $=\dfrac{179}{16}\,\text{m s}^{-1}$
Method: 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 $179/16\,\text{m s}^{-1}$ .

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

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.

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 marks]
Answer
- Time $=1.5\,\text{s}$
- Height gained $=11.025\,\text{m}$
Method: 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 $s=11.025\,\text{m}$ .

Tier 2 · Standard

1. 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.[6 marks]
Answer
- Time of flight $=2.04\,\text{s}$
- Range $=35.3\,\text{m}$
- Greatest height $=5.10\,\text{m}$
Method: The components are $u_x=20\cos30^\circ=10\sqrt3$ and $u_y=20\sin30^\circ=10$ . From vertical displacement $0=10t-4.9t^2$ , the non-zero time is $t=10/4.9=2.0408\,\text{s}$ . The range is $10\sqrt3(2.0408)=35.3\,\text{m}$ . At the top, $0=10^2-2(9.8)h$ , so $h=100/19.6=5.10\,\text{m}$ .

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 marks]
Answer
- $y=5+\dfrac{5}{12}x-\dfrac{5}{144}x^2$
- Horizontal distance $=6(1+\sqrt5)\,\text{m}$
- Impact speed $=\sqrt{269}\,\text{m s}^{-1}$
Method: $x=12t$ 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 $\sqrt{12^2+(5\sqrt5)^2}=\sqrt{269}\,\text{m s}^{-1}$ .