A) 4s
B) 10 s
C) 12 s
D) 8 s
Correct Answer: D
Solution :
Key Idea: Bus starts from rest hence initial velocity is zero. From equation of motion, we have \[s=ut+\frac{1}{2}a{{t}^{2}}\] where, s is displacement, u is initial velocity, a is acceleration and t is time. Since, bus starts from rest, u = 0 \[\therefore \] \[{{s}_{b}}=+\frac{1}{2}\times 1\times {{t}^{2}}\] Man is 48 m behind the bus \[\therefore \] \[{{s}_{m}}=10\times t+\frac{1}{2}\times 0\times {{t}^{2}}\] \[{{s}_{m}}=10t\] When man catches the bus \[{{s}_{m}}={{s}_{b}}+48\] \[10t=\frac{{{t}^{2}}}{2}+48\] \[\Rightarrow \]\[(t-8)(t-12)=0\] Hence, minimum time after which man will catch the bus is 8 s.
You need to login to perform this action.
You will be redirected in
3 sec
