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

Nov 6, 2017

The average speed is $= \frac{29}{3} m {s}^{-} 1$

Explanation:

The speed is the integral of the acceleration

$a \left(t\right) = 2 {t}^{2} - t$

$v \left(t\right) = \int \left(2 {t}^{2} - t\right) \mathrm{dt} = \frac{2}{3} {t}^{3} - \frac{1}{2} {t}^{2} + C$

The initial conditions are

$v = 4 m {s}^{-} 1$ when $t = 0$

Plugging these values in the equation of $v \left(t\right)$

$v \left(0\right) = \frac{2}{3} \cdot 0 - \frac{1}{2} \cdot 0 + C = 4$

Therefore,

$C = 4$

$v \left(t\right) = \frac{2}{3} {t}^{3} - \frac{1}{2} {t}^{2} + 4$

The average speed is

$\left(4 - 0\right) \overline{v} = {\int}_{0}^{4} \left(\frac{2}{3} {t}^{3} - \frac{1}{2} {t}^{2} + 4\right) \mathrm{dt}$

$4 \overline{v} = {\left[\frac{1}{6} {t}^{4} - \frac{1}{6} {t}^{3} + 4 t\right]}_{0}^{4}$

$4 \overline{v} = \left(\frac{128}{3} - 8 + 4\right) - \left(0\right)$

$\overline{v} = \frac{1}{4} \cdot \frac{116}{3} = \frac{29}{3} m {s}^{-} 1$