A car and a bus are travelling along a straight road towards traffic lights. The traffic lights change at time $t = 0$, where $t$ is in seconds. At this instant the car has a speed of $15 ms^{-1}$
. The car then
- decelerates uniformly to rest in $22.5 m$ (as in part (a)),
- waits at the traffic lights for $7$ seconds,
- accelerates uniformly up to $15 ms^{-1}$
- in $5$ seconds, travels at $15 ms^{-1}$ down the road.
When $t = 0$ the car is level with a bus which is travelling at a constant speed of $20 ms^{-1}$ along a bus lane. The bus is not required to stop at the traffic lights and continues at this speed down the road.
The bus has travelled $240 m$ further than the car at the time that the car again reaches $15 ms^{-1}$
At the instant that the car reaches its constant speed of $15 ms^{-1}$, the bus begins to decelerate uniformly at $0.2 ms^{-2}$.
(iii) It takes T seconds for the car to catch up with the bus after the bus begins to decelerate. Show that the car must travel $240 + 20T - 0.1T^2$ in this time. Hence show that $T$ satisfies the equation $T^2 - 50T - 2400 = 0$.
Find the speed of the bus when the car catches up with it.
I have managed to show that the distance that the car has travelled is indeed $240 + 20T - 0.1T^2$, but cannot figure out how to show how it satisfies the last equation and how to find the car's speed. I assume it has to do with finding $v$, but don't know how to go about it. Can anybody help me?
