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

Apr 24, 2016

${v}_{a} = - 3 \text{ unit/s}$

#### Explanation: $v \left(t\right) = \int \left(t - 3\right) \cdot d t \text{ } v \left(t\right) = {t}^{2} / 2 - 3 t$

$x \left(t\right) = \int v \left(t\right) \cdot d t \text{ } x \left(t\right) = \int \left({t}^{2} / 2 - 3 t\right) \cdot d t$

$x \left(t\right) = \frac{1}{2} \cdot {t}^{3} / 3 - 3 {t}^{2} / 2 \text{ } x \left(t\right) = {t}^{3} / 6 - \frac{3 {t}^{2}}{2}$

$\text{for position t=3 -> x(3)=-9 unit}$

$\text{for position t=6 ->x(6)=-18 unit}$

${v}_{a} = \tan \alpha$

${v}_{a} = \frac{- 18 + 9}{6 - 3}$

${v}_{a} = - \frac{9}{3}$

${v}_{a} = - 3 \text{ unit/s}$

Apr 25, 2016

$\text{Average speed"=-3 " units}$
$- v e$ sign indicates that speed is opposite to the defined positive direction of displacement.

#### Explanation:

Given acceleration $a \left(t\right) = t - 3$. We know that $a = x ' '$, therefore, to find distance moved during the period of interest $x = \int \int a \left(t\right) . \mathrm{dt} . \mathrm{dt}$ over the period of interest
(First integral will give us the velocity in terms of time.) Inserting given values.

$x = \int \int \left(t - 3\right) . \mathrm{dt} . \mathrm{dt}$
$= {\int}_{3}^{6} \left({t}^{2} / 2 - 3 t + C\right) . \mathrm{dt}$ .....(1), where $C$ is constant of integration.
We need to evaluate $C$ from initial conditions
Velocity $v \left(t\right) = {t}^{2} / 2 - 3 t + C$
Now $v \left(0\right) = {0}^{2} / 2 - 3 \times 0 + C = 0$, initial condition
or $C = 0$
Inserting in (1)

$x = {\int}_{3}^{6} \left({t}^{2} / 2 - 3 t\right) . \mathrm{dt}$
$x = | {t}^{3} / 6 - 3 {t}^{2} / 2 {|}_{3}^{6}$, constant of integration ignored due to definite integral

$= \left({6}^{3} / 6 - 3 \times {6}^{2} / 2\right) - \left({3}^{3} / 6 - 3 \times {3}^{2} / 2\right)$
$= 36 - 54 - \frac{9}{2} + \frac{27}{2}$
$= - 9$,
We know that $\text{Average speed"=("Total Displacement")/"Time Taken}$
$\text{Average speed} = - \frac{9}{6 - 3}$
$\text{Average speed} = - 3$