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]?

Mar 15, 2018

Given, acceleration =$a = \frac{\mathrm{dv}}{\mathrm{dt}} = 2 - 5 t$

so,$v = 2 t - \frac{5 {t}^{2}}{2} + 12$ (by integration)

Hence, $v = \frac{\mathrm{dx}}{\mathrm{dt}} = 2 t - \frac{5 {t}^{2}}{2} + 12$

so,$x = {t}^{2} - \frac{5}{6} {t}^{3} + 12 t$

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

So,total distance covered =${\left[{t}^{2}\right]}_{0}^{3.23} - \frac{5}{6} {\left[{t}^{3}\right]}_{0}^{3.23} + 12 {\left[t\right]}_{0}^{3.23} + \frac{5}{6} {\left[{t}^{3}\right]}_{3.23}^{4} - {\left[{t}^{2}\right]}_{3.23}^{4} - 12 {\left[t\right]}_{3.23}^{4} = 31.54 m$

So,average velocity = total distance covered/total time taken =$\frac{31.54}{4} = 7.87 m {s}^{-} 1$