# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) =2t^2-3t-2 from t in [2, 4]?

Dec 27, 2015

$15.32 \text{m/s}$

#### Explanation:

Acceleration is a function of time here so the average velocity will be the area under the acceleration - time graph between the limits given.

graph{2x^2-3x-2 [-10, 10, -5, 5]}
$\therefore {v}_{a v e} = {\int}_{2}^{4} a . \mathrm{dt}$

${v}_{a v e} = {\int}_{2}^{4} \left(2 {t}^{2} - 3 t - 2\right) . \mathrm{dt}$

${v}_{a v e} = {\left[\frac{2 {t}^{3}}{3} - \frac{3 {t}^{2}}{2} - 2 t\right]}_{2}^{4}$

v_(ave)=[(2xx4^(3))/(3)-(3xx4^(2))/2-(2xx4)]-[(2xx2^(3)-(3xx2^(2))/2-(2xx2)]

${v}_{a v e} = \left[\frac{128}{3} - 24 - 8\right] - \left[\frac{16}{3} - 6 - 4\right]$

${v}_{a v e} = 10.66 - \left(- 4.66\right)$

${v}_{a v e} = 15.32 \text{m/s}$