Question

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]`?

Answer 1

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`