# A bus is beginning to move with an acceleration of 1 m/s^2. A boy who is 48 m behind the bus starts running at 10 m/s. After what time will the boy be able to catch the bus?

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Let us say when boy catch the bus the bus has run xm from its initial position. So at that time boy runs (48+x)m. Let us say the time taken to catch the bus is t.

For bus

`S = ut+1/2xxaxxt^2`

`x = 1/2xx1xxt^2 = t^2/2`

`x = t^2/2 ----(1)`

For boy

`S = ut+1/2xxaxxt^2`

`48+x = 1/2xx10xxt^2`

`48+x = 5t^2 ---(2)`

(2)-(1)

`48 = 5t^2-t^2/2`

`96 = 9t^2`

`t^2 = sqrt(96/9)`

`t = 3.265s`

*So the boy will catch the bus after 3.265 seconds.*

*Assumptions*

*Both boy and bus start the motion from rest.**Both bus and boy start the motion simultaneously*

**Sources:**