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#

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Answer 1 · 2017-03-22T21:00:16

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

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 · 2017-03-24T06:20:41

#rArr x = 125/3#

#(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#