Forces and Motion
6 marks · NormalA van of mass 1200 kg is moving along a straight road. The driver notices a traffic jam ahead and comes to rest by applying the brakes. The van slows from 18 m/s to 0 m/s in 4.5 s.
State the SI unit for acceleration.
Calculate the acceleration of the van while braking. State whether this acceleration is positive or negative.
Calculate the unbalanced (resultant) force acting on the van while braking.
Explain why the van takes a longer distance to stop if the road is wet.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) metre per second squared (m/s^2)
(b) -4.0 m/s^2
(c) -4800 N
(d) A wet road reduces friction between the tyres and the road, so the braking force is smaller and the van takes longer to stop.
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · States correct SI unit of acceleration: m/s² or metre per second squared.
M1 (b) · 1 mark · Uses correct equation: a = (v-u)/t, includes correct sign for direction.
A1 (b) · 1 mark · Correct value: -4.0 m/s² (accept -4 m/s²) with unit and sign.
M2 (c) · 1 mark · Applies F = m × a using own value for a. (follow-through)
A2 (c) · 1 mark · Correct value for resultant force: -4800 N (accept sign or 'opposite to motion'). (follow-through)
B2 (d) · 1 mark · Mentions friction and links decrease in friction to longer stopping distance.
Worked solution
a) The SI unit for acceleration is metre per second squared (m/s²).
b) Initial velocity u = 18 m/s, final velocity v = 0 m/s, time t = 4.5 s
This acceleration is negative (deceleration), as the van is slowing down.
c) Using F = m × a: F = 1200 kg × (-4.0 m/s²) = -4800 N
d) On a wet road there is less friction between tyres and the road, so the braking force is smaller. This means the van takes a longer distance to stop.