# What is the average speed, on t in [0,5], of an object that is moving at 12 m/s at t=0 and accelerates at a rate of a(t) =5t^2-3t on t in [0,4]?

Feb 12, 2018

Given:
Initial velocity of the object $v 0 = 12 \frac{m}{s}$ at $t = 0$
To determine the average speed for$t \left(0 , 5\right)$
The acceleration is given by
$a \left(t\right) = 5 {t}^{2} - 3 t$
We know that
$a \left(t\right) = \frac{\mathrm{dv}}{\mathrm{dt}}$
Separating variables,
$\mathrm{dv} = a \left(t\right) \mathrm{dt}$
Substituting for $a \left(t\right)$
$\mathrm{dv} = \left(5 {t}^{2} - 3 t\right) \mathrm{dt}$
Integrating both sides
$\int \mathrm{dv} = \int \left(5 {t}^{2} - 3 t\right) \mathrm{dt}$
$v = \frac{5}{3} {t}^{3} - \frac{3}{2} {t}^{2} + v 0$

At $t = 0 , v 0 = 12 \frac{m}{s}$
Substituting
$12 = \frac{5}{3} {\left(0\right)}^{3} - \frac{3}{2} {\left(0\right)}^{2} + v 0$
$12 = 0 - 0 + v 0$
$v 0 = 12$
Thus, the expression for the velocity given by
$v \left(t\right) = \frac{5}{3} {t}^{3} - \frac{3}{2} {t}^{2} + 12$
Velocity at 4 sec is
$v \left(4\right) = \frac{5}{3} {\left(4\right)}^{3} - \frac{3}{2} {\left(4\right)}^{2} + 12$
$v \left(4\right) = 106.667 - 37.5 + 12$
$v \left(4\right) = 81.167$m/s
Average velocity for the interval is
Distance travelled is obtained by integrating the expression of velocity
$v = \frac{\mathrm{ds}}{\mathrm{dt}}$
Separating variables,
$\mathrm{ds} = v \left(t\right) \mathrm{dt}$
Substituting for $v \left(t\right)$
$\mathrm{dv} = \frac{5}{3} {t}^{3} - \frac{3}{2} {t}^{2} + 12 \mathrm{dt}$

Integrating both sides
$\int \mathrm{dv} = \int \left(\frac{5}{3} {t}^{3} - \frac{3}{2} {t}^{2} + 12\right) \mathrm{dt}$
$s \left(t\right) = \frac{5}{12} {t}^{4} - \frac{1}{2} {t}^{3} + 12 t + s 0$
Assuming the object starts from rest, s0=0
$s \left(t\right) = \frac{5}{12} {t}^{4} - \frac{1}{2} {t}^{3} + 12 t + s 0$
At t-4s, the position is
$s \left(4\right) = \frac{5}{12} {\left(4\right)}^{4} - \frac{1}{2} {\left(4\right)}^{3} + 12 t + 0$
$s \left(4\right) = 122.667$

Total distance travelled in the interval of 4 seconds is $122.667$m
Total time elapsed is 4 seconds

average speed = distance /time
$= \frac{122.667}{4} = 30.667 \frac{m}{s}$