W2.2 Traffic accident report
You have taken a summer job with an insurance company and are helping to investigate an accident. Having admitted taking Physics, you are duly dispatched to the scene. Here you find a road that runs straight down a hill inclined at 10 degrees to the horizontal. At the bottom of the hills, the road flattens out into a parking area, overlooking a cliff.
THe cliff has a vertical drop to horizontal ground below (you don’t know the height, but it is a long way down!), where a car lies wreck 30 feet away from the base of the cliff. A witness claims that the car was parked on the hill and inexplicably began coasting down the road, taking about 3 seconds to get to the bottom.
Your boss, who has accompanied you to the scene, drops a stone over the edge of the cliff and times that it takes 5 seconds until he hears it land at the base of the cliff below.
Your boss suspects foul play. Based on the witness statement and other factors in the case, what do you think?
Hint
Reveal

Hint
You need to break the motion down into stages, and work out a theory to test if the information is correct or not.Solution
Reveal

Solution
I’ll start this from the ’P’ part of our FDPEE strategy, assuming that you can do the ’focus’ and ’describe’ parts yourself.
Our strategy to solve the problem is as follows:
- Use the information on the dropped stone to estimate the height of the cliff.
- Based on the witness, calculate a speed for the car at the bottom of the hill
- Calculate how far from the base of the cliff this ’projectile’ would land.
- Compare with where the wreck is found, and draw conclusions
1. Dropped stone
You boss claims it takes 5s to land; we can use y=-1/2gt2 to calculate the vertical drop (y), using the simplified version as the expression as the stone is released from rest. This gives y=123m
(A note on units: people often measure things in feet, yards etc. Although in principle we could do all our calculations in non-SI units (as in Speed kills), sometimes it is easier to keep things in SI units. We know that g=9.8ms-2, but do you know what value it has in feet/s2? )
2. Rolling car
If the car rolls for 3s down the sloping hill, we should be able to calculate the speed at the bottom. Assuming no friction, this will also be the speed it goes over the edge.
What is accelerating the car down the slope is the component of g acting down the slope (=gsinθ)
We can estimate the speed at the bottom of the slope using v=u+at, or simply v=at as the car presumably rolls from rest. Putting in the numbers gives us v=5ms-1.
3. Projectile car.
The car goes over the cliff at this speed. We can now calculate howe far away it will land from the base of the cliff. Recall the argument made in S1.3 that we can treat the x and y components of the motion separately. The x-motion is governed by x=vx0×t. The t here is the same for the stone that your boss dropped (in fact it is the same for anything falling off the cliff!).
So our calculated distance is x=25m. This is about 80 feet, far greater than the 30 suggested in the question.
4. Verdict!
So what can you conclude. Well, you can say that the car went over the cliff at a much slower speed than that accquired rolling down the hill. It sounds a lot like an insurance job, in which the car was pushed over the edge at a slow speed.
Just occasionally we get asked to do this sort of thing for real: accident investigations (or solicitors) have emailed the School in the past to ask for ’expert analysis’ of a crash scene. The scenario is usually more complex than this one, but the underlying physics is the same.
More crash mayhem in a few weeks time in Q5.9