3.1 Speed, velocity and motion graphs — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

Two forces act on a trolley: 45 N forwards and 20 N backwards. Calculate the resultant force.
Calculate the energy in the kinetic store of a 3.0 kg object moving at 4.0 m/s.

Answers

The forces act in opposite directions, so subtract: $45 - 20 = 25\,\text{N}$ forwards.
$E_k = \tfrac{1}{2}mv^2 = 0.5 \times 3.0 \times 4.0^2 = 24\,\text{J}$ .

If question 1 went wrong, look again at Section 2.1: forces in opposite directions subtract. If question 2 went wrong, remember to square the speed before you multiply.

Practise with help

1. A train travels 450 m in 30 s. Calculate its average speed.

Hint

Average speed is the total distance divided by the total time.

Next step

Write $v = \dfrac{s}{t}$ , then divide 450 m by 30 s.

Full working

\begin{aligned} v &= \frac{s}{t} \\ &= \frac{450}{30} \\ &= 15\,\text{m/s} \end{aligned}

2. A walker moves at a steady 1.4 m/s for 300 s. Calculate the distance she travels.

Hint

The distance travelled is the speed multiplied by the time.

Next step

Write $s = vt$ , then multiply 1.4 m/s by 300 s.

Full working

\begin{aligned} s &= v t \\ &= 1.4 \times 300 \\ &= 420\,\text{m} \\ &= 4.2 \times 10^2\,\text{m} \end{aligned}

3. A car speeds up from 6.0 m/s to 18 m/s in 4.0 s. Calculate its acceleration.

Hint

Acceleration uses the change in velocity, not the final velocity.

Next step

Work out the change first: $18 - 6.0$ . Then divide by 4.0 s.

Full working

\begin{aligned} a &= \frac{\Delta v}{t} \\ &= \frac{18 - 6.0}{4.0} \\ &= \frac{12}{4.0} \\ &= 3.0\,\text{m/s}^2 \end{aligned}

4. A velocity-time graph is a straight line through $(0\,\text{s}, 0\,\text{m/s})$ , $(4.0\,\text{s}, 8.0\,\text{m/s})$ and $(8.0\,\text{s}, 16\,\text{m/s})$ . Calculate the acceleration from the gradient.

Hint

The acceleration is the gradient of a velocity-time graph.

Next step

Use the two end points: change in velocity is $16 - 0$ , and change in time is $8.0 - 0$ .

Full working

\begin{aligned} a &= \frac{\text{change in velocity}}{\text{change in time}} \\ &= \frac{16 - 0}{8.0 - 0} \\ &= 2.0\,\text{m/s}^2 \end{aligned}

5. A van travels at 16 m/s. The driver brakes with a steady deceleration of $4.0\,\text{m/s}^2$ until the van stops. Calculate the distance the van travels while braking.

Hint

The van starts at 16 m/s and ends at rest, so the final velocity is zero. The acceleration is negative, because the van slows down.

Next step

Write $v^2 - u^2 = 2as$ , then put in $v = 0$ , $u = 16$ and $a = -4.0$ .

Full working

\begin{aligned} v^2 - u^2 &= 2 a s \\ 0^2 - 16^2 &= 2 \times (-4.0) \times s \\ -256 &= -8.0 s \\ s &= \frac{256}{8.0} \\ &= 32\,\text{m} \end{aligned}

6. A skydiver jumps from a plane. Explain, in terms of forces, why the skydiver reaches a terminal velocity, and why the velocity then stays constant.

Hint

One force pulls the skydiver down, and another force from the air pushes up. Say what happens to the upward force as the skydiver gets faster.

Next step

Say what the resultant force is when the two forces become equal, and what a zero resultant force does to the velocity.

Full working

At first the weight of the skydiver is larger than the drag force from the air, so the resultant force is downwards and the skydiver accelerates. As the skydiver gets faster, the drag force increases. When the drag force equals the weight, the resultant force is zero. A zero resultant force means the velocity stops changing, so the skydiver falls at a steady terminal velocity. This answer is a model answer for you to compare with your own wording.