3.1 Kinematics 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

Write $2.5 \times 10^4\,\text{m}$ in kilometres.
A force of 60 N acts at 30° to the horizontal. Find its horizontal component.
Work out the area of a trapezium with parallel sides 6.0 cm and 14 cm, and height 5.0 cm.

Answers

$2.5 \times 10^4\,\text{m} = 25\,\text{km}$ .
$60\cos 30^\circ = 52.0\,\text{N}$ .
$\tfrac{1}{2}(6.0 + 14) \times 5.0 = 50\,\text{cm}^2$ .

If question 1 goes wrong, look again at the prefix table in Section 1.1. If question 2 goes wrong, the angle is measured from the horizontal, so that component uses the cosine. If question 3 goes wrong, the area of a trapezium is half the sum of the parallel sides, multiplied by the height.

Practise with help

1. A cyclist accelerates uniformly from rest to $8.0\,\text{m}\,\text{s}^{-1}$ in 5.0 s. Find the acceleration and the distance travelled.

Hint

The cyclist starts from rest, so the initial velocity is zero. The acceleration is the change in velocity each second.

Next step

Use $a = \dfrac{v - u}{t}$ with $u = 0$ , then use $s = ut + \tfrac{1}{2}at^2$ .

Full working

\begin{aligned} a &= \frac{8.0 - 0}{5.0} = 1.6\,\text{m}\,\text{s}^{-2} \\ s &= ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2} \times 1.6 \times 5.0^2 \\ &= 20\,\text{m} \end{aligned}

The distance is 20 m. The same distance comes from the area under the velocity–time graph, which is a triangle of base 5.0 s and height 8.0 m s⁻¹.

2. A car slows uniformly from $24\,\text{m}\,\text{s}^{-1}$ to rest in 6.0 s. Find its deceleration and the distance it travels while braking.

Hint

Take the direction of travel as positive. The final velocity is zero, so the change in velocity is negative.

Next step

Find $a$ from $a = \dfrac{v - u}{t}$ , then use $s = \dfrac{(u + v)t}{2}$ .

Full working

\begin{aligned} a &= \frac{0 - 24}{6.0} = -4.0\,\text{m}\,\text{s}^{-2} \\ s &= \frac{(u + v)t}{2} = \frac{(24 + 0) \times 6.0}{2} = 72\,\text{m} \end{aligned}

The car slows at $4.0\,\text{m}\,\text{s}^{-2}$ and travels 72 m before it stops.

3. A van's velocity–time graph rises in a straight line from 0 to $20\,\text{m}\,\text{s}^{-1}$ in 8.0 s, stays at $20\,\text{m}\,\text{s}^{-1}$ for 12 s, then falls in a straight line to rest in 4.0 s. Find the total displacement and the deceleration in the last stage.

Hint

The displacement is the area under the graph. Split the area into a triangle, a rectangle and a triangle.

Next step

The deceleration is the gradient of the last section, which falls from 20 to 0 over 4.0 s.

Full working

\begin{aligned} s &= \tfrac{1}{2}(8.0)(20) + (12)(20) + \tfrac{1}{2}(4.0)(20) \\ &= 80 + 240 + 40 = 360\,\text{m} \\ a &= \frac{0 - 20}{4.0} = -5.0\,\text{m}\,\text{s}^{-2} \end{aligned}

The displacement is 360 m, and the van slows at $5.0\,\text{m}\,\text{s}^{-2}$ in the last stage. A falling line on a velocity–time graph is a negative acceleration.

4. A stone is dropped from rest down a well and falls 20 m. Find the time it takes to fall and its speed at the bottom. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

Take downwards as positive. The stone starts from rest, and the acceleration is $g$ .

Next step

Use $s = ut + \tfrac{1}{2}at^2$ with $u = 0$ and $a = g$ to find $t$ , then use $v = u + at$ .

Full working

\begin{aligned} 20 &= \tfrac{1}{2} \times 9.81 \times t^2 \\ t &= \sqrt{\frac{40}{9.81}} = 2.0\,\text{s} \\ v &= 9.81 \times 2.0 = 19.6\,\text{m}\,\text{s}^{-1} \end{aligned}

The stone falls for 2.0 s and reaches about $20\,\text{m}\,\text{s}^{-1}$ . Free fall is much faster than most people expect.

5. The chapter problem. The roller-coaster car is released from rest down a straight section of track at 60° to the horizontal. It descends 45 m vertically. Find the acceleration of the car down the slope, the length of the slope, and its speed at the bottom.

Hint

The only force along the slope is the component of the weight, so $a = g\sin 60^\circ$ . The vertical drop is the opposite side of a right-angled triangle whose hypotenuse is the slope.

Next step

Find $a = 9.81\sin 60^\circ = 8.50\,\text{m}\,\text{s}^{-2}$ and $s = \dfrac{45}{\sin 60^\circ} = 52.0\,\text{m}$ . Then use $v^2 = u^2 + 2as$ from rest.

Full working

\begin{aligned} a &= 9.81\sin 60^\circ = 8.50\,\text{m}\,\text{s}^{-2} \\ s &= \frac{45}{\sin 60^\circ} = 52.0\,\text{m} \\ v^2 &= 0 + 2 \times 8.50 \times 52.0 = 884 \\ v &= 29.7\,\text{m}\,\text{s}^{-1} \end{aligned}

The car reaches $29.7\,\text{m}\,\text{s}^{-1}$ at the bottom. The same answer comes from energy: the speed depends on the vertical drop of 45 m, not on the angle of the slope.

6. A train leaves a station and reaches $12\,\text{m}\,\text{s}^{-1}$ in 20 s. It travels at $12\,\text{m}\,\text{s}^{-1}$ for 60 s, then slows uniformly to rest in 10 s. A student tries to find the total distance with a single equation of motion. Explain why that fails, and describe how to find the distance correctly.

Hint

Look at the acceleration in each stage. Does it stay the same?

Next step

Name the three stages, and say what is constant in each. For the distance, name the graph whose area gives displacement.

Full working

The equations of motion apply only while the acceleration is constant. This journey has three stages: the train speeds up, travels at a constant velocity, then slows down. The acceleration is different in each stage, so one equation cannot describe the whole journey.

Split the journey into the three stages and find the area under the velocity–time graph. The areas are

\begin{aligned} \tfrac{1}{2}(20)(12) + (60)(12) + \tfrac{1}{2}(10)(12) &= 120 + 720 + 60 \\ &= 900\,\text{m} \end{aligned}

The method is to draw the velocity–time graph first, then find its area in stages.