M1 question

A car is travelling along a straight horizontal road with constant acceleration. The car passes over three consecutive points A, B and C where AB=100m and BC=300m. The speed of the car at B is 14m s−1 and the speed of the car at C is 20m s−1. Find (a) the acceleration of the car, (b) the time taken for the car to travel from A to C.

For part a the acceleration is 0.34 which is correct

for part b) What I done was use a = 0.34, v = 20, s = 400 and used the formula s = vt - (at^2)/2 to get a quadratic of t which I got to be:

0.17t^2 -20t + 400 = 0

and I use this to find the two values of t which I get to be 92.1... and 25.54... In the back of the book theyve specified the answer to be 25.5, but could anyone explain why?

Reply 1

Augmented hippo

The answer of 92.1... would be if the car was initially moving at a speed in opposite direction to AB. If you work out the speed when s=0 using

v^2=u^2+2as

you get

u^2=128

. Because the question says it passes 3 consecutive points (not A twice then B and C) you should only use the positive route. In an exam I think either the wording would be clearer or they would allow both answers.

Reply 2

dantheman1261

Basically, the reason there are two results is that the information used in that particular formula, s = vt - (at^2)/2, is not quite 'complete' as the initial velocity can take two different values.

From the information given, you can work out the initial velocity of the car. However, visualize what would happen if the car started off with the same initial velocity in the opposite direction.

Note that the acceleration is still in the same direction as before, but now opposite to the direction of the initial motion. So the car is moving from A, in the opposite direction to B, but slowing down. Then it stops, and accelerates back towards A. When it reaches A, it will be travelling at the same initial speed, but this time towards B.

So if the car starts off travelling in the opposite direction, it still ends up at C, just after much longer. The final velocity is still 20m/s, the final distance is still 400m (though there was a lot of other travelling backwards and forwards which cancelled out), and the acceleration was still 0.34 in the direction towards C.

Hopefully that wasn't too long an explanation - but all the information you put into the formula can generate two different results depending on the initial direction of movement at A (which isn't accounted for in this particular formula). So you just have to compare the two results and see which one is reasonable! Usually one result is negative, which helps to decide.

Reply 3

Lbb22

Can you explain how you worked out the acceleration please :smile:

Reply 4

davros

Original post by Lbb22

Can you explain how you worked out the acceleration please :smile:

After 8 years the OP might not be around :biggrin:

You have information in the question about the motion between B and C. Since you are told that acceleration is constant you can use SUVAT. Which equation do you think is most helpful for the part of the journey from B to C?

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