Question

A man is standing 40 m behind the bus. Bus starts with 1m per second square constant acceleration and also at the same instant the man starts moving with constant speed 9mper sec. Find the time taken by man to catch the bus

Answer 1

speed of man=9m/s. acceleration according to man will be negative=-1m/s^2. distance=40m.
now s=ut+1/2at^2
40=9t+1/2(-1)t^2
t^2-18t+80=0
(t-8)(t-10)=0
t=8,10

Answer 2

A man is standing 40m behind the bus. Bus starts with 1 m/s² constant acceleration and man starts moving at the same instant with speed speed 9m/s.

here we have to use relative concepts .
relative velocity of man with respect to bus , $v_{mb}=v_m-v_b=v_m-0=v_m=9m/s$

relative acceleration of man with respect to bus, $a_{mb}=a_m-a_b=0-a_b=-1m/s^2$

relative displacement of man with respect to the bus, $S_{mb}=S_m-S_b=40-0=40m$

now, use formula, $S_{mb}=v_{mb}t+\frac{1}{2}a_{mb}t^2$

40 = 9t - 1/2t²
80 = 18t - t²
t² - 18t + 80 = 0
t² - 10t - 8t + 80 = 0
(t - 10)(t - 8) = 0
t = 10, 8

hence, man catches the bus after 8 sec and after 10sec.