Mechanics unit 1

A car is travelling along a straight horizontal road. The car takes 120 s to travel between
two sets of traffic lights which are 2145 m apart. The car starts from rest at the first set of
traffic lights and moves with constant acceleration for 30 s until its speed is 22 m s–1. The
car maintains this speed for T seconds. The car then moves with constant deceleration,
coming to rest at the second set of traffic lights.
(a) Sketch, in the space below, a speed-time graph for the motion of the car between the
two sets of traffic lights.
(2)
(b) Find the value of T.
(3)
A motorcycle leaves the first set of traffic lights 10 s after the car has left the first set of
traffic lights. The motorcycle moves from rest with constant acceleration, a m s–2, and
passes the car at the point A which is 990 m from the first set of traffic lights. When the
motorcycle passes the car, the car is moving with speed 22 m s–1.
(c) Find the time it takes for the motorcycle to move from the first set of traffic lights to
the point A.
(4)
(d) Find the value of a.
I dont understand part c, I've looked at the mark scheme and it doesn;t help so if anyone can explain please, thanks

A motorcycle leaves the first set of traffic lights 10 s after the car has left the first set of
traffic lights. The motorcycle moves from rest with constant acceleration, a m s–2, and
passes the car at the point A which is 990 m from the first set of traffic lights. When the
motorcycle passes the car, the car is moving with speed 22 m s–1.
(c) Find the time it takes for the motorcycle to move from the first set of traffic lights to
the point A.

Right, I worked out that the time taken by the car is 60, then I subtracted 10 seconds seeing as the motorcycle started 10 seconds after the car, however I don't see the sense in what I've done, it is the correct answer however doing it this way, it's assuming that the motorcycle accelerates until 22m/s doesn't it? And thats not explicitly stated?
thanks again

If you quote people it makes it easier for them to notice that you've replied

But yes, it is making that assumption and it's stated in the question:

The motorcycle moves from rest with constant acceleration, a m s–2

which means it does accelerate until 22m/s and even beyond.

Right, I'm still not sure...
I have done area of triangle for initial accel of car to find the diastance moved during that accelerating period= 0.5(30)*22=330
Then minus that from the overall area=990-330=660 then the time taken to move this distance is 660/22=30 30+30=60
60-10=50
What I don't understand is why we are using the cars values, I understand that they meet at the same point, however I don't understand that the motorcycle would always be 10 seconds behind the car bcos the car changes speed and then has a constant speed so surely the time difference would change?

Hmm. I'm finding this a bit difficult to explain.

It may help if you haven''t already to draw the distance-time line for the motorcycle on the one you've drawn already.

They meet at a time t. But the clock starts ticking from 0 - the motorcycle does nothing for 10 seconds while the car starts moving. Yes, the motorcycle starts 10 seconds after the car but it doesn't mean that distance wise they're always 10 seconds apart.

The motorcycle keeps closing the gap until they meet at time t. The car takes t seconds to get to that point, whereas the motorcycle takes t-10 seconds, because of the start. Hence you're using the car values because you know the motorcycle did the distance 10 seconds faster than the car did, and you have enough information for the car.

Let me know if that hasn't cleared things up and I'll try and think of something a bit easier to understand.