15.2 The language of kinematics — Practise with help

Work each one first, closed book. Before you calculate, choose a positive direction and write down the sign of every vector you will use.

Warm-up: remember

Convert $36$ km/h to m/s.
Work out $(-3) + 7$ and $4 - 9$ .
A car travels $180$ m in $9$ s. Work out its average speed.

Answers

$36 \times \dfrac{5}{18} = 10$ m/s.
$(-3) + 7 = 4$ , and $4 - 9 = -5$ .
Average speed $= \dfrac{180}{9} = 20$ m/s.

If question 2 was hard, revise adding and subtracting negative numbers before you continue. Every sign in this section depends on that skill.

Practise with help

1. A walker walks $300$ m north, then turns and walks $100$ m south. Find the distance travelled and the displacement. Take north as positive.

Hint

Distance travelled adds the lengths of the two stages. Displacement uses the signs of the two stages.

Next step

Add $300$ and $100$ for the distance. For the displacement, add $+300$ and $-100$ .

Full working

Distance travelled $= 300 + 100 = 400$ m.

Displacement $= (+300) + (-100) = +200$ m, which is $200$ m north.

The walker doubled back, so the distance is larger than the size of the displacement.

2. A runner completes one lap of a $400$ m track in $80$ s. Find the distance travelled, the displacement and the average speed.

Hint

The finish of a lap is the starting point. Think about the change in position over the whole lap.

Next step

The distance is the whole lap. For the displacement, compare the finish with the start.

Full working

Distance travelled $= 400$ m. The runner finishes where they started, so the displacement is $0$ m.

Average speed $= \dfrac{400}{80} = 5$ m/s.

The average velocity is $0$ m/s, because the displacement is zero.

3. A car travels $150$ m due east in $10$ s, then continues $50$ m due east in $5$ s. Find the average speed and the average velocity over the whole journey.

Hint

The car does not turn around, so the distance travelled and the size of the displacement are equal for this journey.

Next step

Add the two distances and the two times. Then divide.

Full working

Total distance $= 150 + 50 = 200$ m, and total time $= 10 + 5 = 15$ s.

\text{average speed} = \frac{200}{15} = 13\frac{1}{3} \text{ m/s}.

The displacement is $200$ m due east, so the average velocity is also $13\frac{1}{3}$ m/s due east.

4. A cyclist speeds up from $5$ m/s to $20$ m/s in $3$ s. Find the acceleration.

Hint

Acceleration is the change in velocity divided by the time. Subtract the starting velocity from the finishing velocity.

Next step

The change in velocity is $20 - 5 = 15$ m/s. Divide by $3$ s.

Full working

a = \frac{v - u}{t} = \frac{20 - 5}{3} = \frac{15}{3} = 5 \text{ m/s}^2.

The acceleration is positive, and the cyclist is speeding up.

5. A ball is thrown vertically upwards at $9.8$ m/s. It returns to the thrower's hand $2.0$ s later, moving downwards at $9.8$ m/s. Taking upwards as positive, find the acceleration of the ball.

Hint

The starting velocity is $+9.8$ m/s and the finishing velocity is $-9.8$ m/s. Use the acceleration formula with those signs.

Next step

The change in velocity is $(-9.8) - (+9.8) = -19.6$ m/s. Divide by $2.0$ s.

Full working

a = \frac{v - u}{t} = \frac{-9.8 - 9.8}{2.0} = \frac{-19.6}{2.0} = -9.8 \text{ m/s}^2.

The acceleration is $-9.8$ m/s $^2$ throughout the flight. The negative sign means gravity points downwards, opposite to the positive direction.

6. A cyclist rides east at $6$ m/s, turns around and later rides west at $6$ m/s. Taking east as positive, write down the velocity before and after the turn. Has the speed changed?

Hint

Velocity needs a direction, and the sign carries it. Speed is the size of the velocity.

Next step

Before the turn, the velocity is in the positive direction. After the turn, it is in the negative direction with the same size.

Full working

Before the turn the velocity is $+6$ m/s, or $6$ m/s east. After the turn the velocity is $-6$ m/s, or $6$ m/s west. The speed is $6$ m/s in both cases, so the speed has not changed but the velocity has.

7. A car has velocity $-15$ m/s and acceleration $+3$ m/s $^2$ . Is the car speeding up or slowing down? Give a reason.

Hint

Compare the directions of the velocity and the acceleration. The velocity is negative, so the car moves in the negative direction.

Next step

The acceleration is positive, which is opposite to the velocity. Decide what happens to the size of the velocity when the two point in opposite directions.

Full working

The velocity is $-15$ m/s, so the car moves in the negative direction. The acceleration is $+3$ m/s $^2$ , which points in the positive direction, opposite to the motion. The acceleration reduces the size of the velocity, so the car is slowing down. After $5$ s the velocity would be $-15 + 3 \times 5 = 0$ m/s.