If the bus is travelling from rest, starting 50m away from the man, with uniform acceleration
a, then its SUVAT equation for displacement is given by:
sbus=50+0t+21at2.
Since the man is running at a constant speed, lets say it's
u, then his displacement is given by:
sman=0+ut+210t2.
These simplify to be:
sbus=50+21at2sman=ut.
30 seconds later, the man catches up to the bus. At this point, they both have the same displacement. Hence,
sbus=sman must hold for
t=30. This yields you an equation in
a and
u.
You need another equation in
a and
u in order to solve these two simultaneously for acceleration of the bus and the speed of the man.
This other equation is not so obvious, but we know that the man starts running at some speed u and the bus has increasing speed starting at 0.
- While the speed of the bus is less than u, the man is catching up to the bus.
- When the speed of the bus is equal to u, the man is no longer catching up to the bus. They maintain a constant separation distance for a moment.
- Once the speed of the bus is greater than u, the distance between the man and the bus is increasing, as the bus is driving away.
We are told that the man *just* catches the bus. This means that the speed of the bus is < u when time is
0≤t<30, and when
t=30 they must obtain the same speed at the moment the man catches up to it.
This yields our second equation, speed of the bus at t=30 is equal to u. I.e.
30a=u.
EDIT: Beaten to it