# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) =12-2t from t in [3, 6]?

Nov 8, 2016

Given
Acceleration $a \left(t\right) = 12 - 2 t$

$\implies \frac{d \left(v \left(t\right)\right)}{\mathrm{dt}} = 12 - 2 t$

$\implies \int \left(d \left(v \left(t\right)\right)\right) = \int \left(12 - 2 t\right) \mathrm{dt}$

$\implies v \left(t\right) = 12 t - 2 \times {t}^{2} / 2 + c$

$\implies v \left(t\right) = 12 t - {t}^{2} + c$

where t = integration constant

Now by the given condition

at t=0, v(0)=0

So $v \left(0\right) = 0 = 12 \times 0 - {0}^{2} + c \implies c = 0$

Hence

$\implies v \left(t\right) = 12 t - {t}^{2}$

$\implies \frac{d \left(s \left(t\right)\right)}{\mathrm{dt}} = 12 t - {t}^{2}$

$\implies {\int}_{3}^{6} \left(d \left(s \left(t\right)\right)\right) = {\int}_{3}^{6} \left(12 t - {t}^{2}\right)$

$\implies s \left(6\right) - s \left(3\right) = {\left[6 {t}^{2} - {t}^{3} / 3\right]}_{3}^{6}$

$= 6 \cdot {6}^{2} - {6}^{3} / 3 - 6 \cdot {3}^{2} + {3}^{3} / 3$

$= 216 - 72 - 54 + 9 = 99$

Average speed in time [3,6] is

$= \frac{s \left(6\right) - s \left(3\right)}{6 - 3} = \frac{99}{3} = 33 u n i t$