al physic question

(a) let t be the time required.
distance covered by man in time t = 5t
distance covered by the bus in time t = (1/2).(0.5).t^2
hence, 5t = 24 + (1/2)(0.5)t^2
solve for t gives t = 8 s ot 12 s
hence, the man first reaches the rear of the bus at 8 s, and then at 12 s when he has fell behind the bus.
(b) Velocity of bus at 8s = 0.5 x 8 m/s = 4 m/s
Assume the man can reach the front of the bus in t' sec after he has reached the rear of the bus.
Distance covered by him = 5t'
Distance covered by the bus = 4t' + (1/2)(0.5)t'^2
hence, 5t' = (4t' + t'^2/4) + 8
i.e. 20t' = 16t' + t'^2 + 32
t'^2 - 4t' +32 = 0
since solving t' gives no real solution, the man cannot reach the front of the bus.

Let t be the time required to reach the bus by the man.

s be the displacement of the rear of the bus, then s + 24 is the displacement of the man.

For the man, s + 24 = vt

s + 24 = 5t ... (1)

For the bus, s = 1/2 at2

s = 1/2 (0.5)t2 ... (2)

Solving (1) and (2), t = 12 or 8

Take the smallest value, t = 8 s

2009-09-27 15:50:14 補充：
b. To reach the front of the bus, the displacement of the man = s + 24 + 8 = s + 32

For the man, s + 32 = 5t ... (3)

For the bus, s = 1/2 at2

s = 1/2 (0.5)t2 ... (4)

There is no such t to satisfy (3) and (4)

So, the man cannot reach the front of the bus.