What is the average speed of an object that is moving at 12 m/s at t=0 and accelerates at a rate of a(t) =2-5t on t in [0,4]?

1 Answer
Mar 15, 2018

Given, acceleration =a=(dv)/(dt)=2-5t

so,v=2t -(5t^2)/2 +12 (by integration)

Hence, v=(dx)/(dt)=2t-(5t^2)/2 +12

so,x=t^2 -5/6 t^3 +12t

Putting,x=0 we get, t=0,3.23

So,total distance covered = [t^2]_0^(3.23) -5/6 [t^3]_0^3.23 +12[t]_0^3.23+5/6[t^3]_3.23^4 -[t^2]_3.23^4 -12[t]_3.23^4=31.54m

So,average velocity = total distance covered/total time taken =31.54/4=7.87 ms^-1