physics problem

a) After t seconds, the distance travelled by the bus is

0.5*a*t^2 = 0.5*(0.5)*t^2 = 0.25*t^2

After t seconds, the distance travelled by the man is

5*t

Since the distance between the man and the bus is 24m, therefore

0.25*t^2 +24 = 5*t ------------------------eqn(1)

t^2 -20*t +96 = 0

(t-8)(t-12) = 0

t = 8 or t =12

Therefore, after 8s and 12s, the man will reach the rear of the bus.

b) If the man had to run to the front of the bus, the distance between the front and the man will be

24 + 8 = 32m

From eqn(1), replace 24 with 32, you have

0.25*t^2 +32 = 5*t

t^2 - 20t + 128 = 0 ----------------------eqn(2)

since b^2-4ac = 20^2 - 4*128 = -112 (<0)

eqn(2) will have complex roots.

Therefore it is NOT possible for the man to reach the front of the bus.

(a)
The distance travelled by the bus when the man takes to reach the rear of the bus is

s = ut + (1/2)at^2

since u = 0,

s = (1/2)at^2

s = (1/2)(0.5)t^2

s = (1/4)t^2

The distance travelled by the man when he takes to reach the rear of bus is

s' = v x t

s' = 5t

since the man reach the rear of bus,

s' - 24 = s

5t - 24 = (1/4)t^2

(1/4)t^2 - 5t + 24 = 0

t^2 - 20t + 96 = 0

t = 8 s or 12 s

(b)
similar to (a)

distance travelled by bus,
(1/4)t^2

distance travelled by man,
s' = 5t

since the man need to reach the front of the bus,

s' - (24 + 8) = s

5t - 32 = (1/4)t^2

t^2 - 20t + 128 = 0

since detla < 0 (i.e. 20^2 - 4 x 1 x 128 = -112 < 0)

so there is no real solution.

The man cannot reach the front of bus.