Question

If acceleration is increasing at the constant rate of `2 m/s^2`. Find the distance travelled in `5 sec`? If initial velocity and acceleration both were zero. (A) `75m` (B) `100m` (C) `125/3 m`

Answer 1

After 5 seconds the object will have reached a speed of:
`v=5sxx2m//s^2=10m//s`

Explanation:

Since the speed goes up from 0 to 10 m/s at a constant rate, the average speed in those 5 seconds will be:

`barv=(0+10)/2m//s=5m//s`

Since distance=(average speed) times (time):

Distance `s=barvxxt=5m/cancelsxx5cancels=25m`

I'm afraid that's not one of the choices you had... but I'm pretty sure it's correct.

Answer 2

`rArr x = 125/3`

Explanation:

`(da)/dt=2`

`rArr da= 2dt`

`rArr int_0^a da= 2 int_0^tdt`

`rArr a=2t`

`rArr (dv)/dt=2t`

`rArr (dv)=2t dt`

`rArr int_0^v (dv)=2 int_0^t *dt`

`rArrv= 2t^2/2`

`rArrv= t^2`

`rArr (dx)/dt= t^2`

`rArr dx= t^2 * dt`

`rArr int_0^x dx= int_0^5 t^2 * dt`

`rArr x = 5^3/3`

`rArr x = 125/3`