# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) = 36-t^2 from t in [0, 5]?

Mar 5, 2016

Average speed $= 79.58 \dot{3} \textrm{u n i t s}$

#### Explanation:

We know that for variable acceleration case

Average velocity ${v}_{\textrm{a v e r a \ge}} = \frac{x \left(t\right) - {x}_{0}}{t}$ .......(1)

Given acceleration $a \left(t\right) = 36 - {t}^{2}$ and at $t = 0$ object is still.
$\implies$ ${v}_{0} = 0$, time $t \in \left[0 , 5\right]$
Now general expressions for
velocity $v \left(t\right) = \frac{\mathrm{dx}}{\mathrm{dt}} = {v}_{0} + {\int}_{0}^{t} a \left(t\right) \cdot \mathrm{dt}$ .......(2)
and Position $x \left(t\right) = {x}_{0} + {\int}_{0}^{t} v \left(t\right) \cdot \mathrm{dt}$ ......(3)
where ${x}_{0} \mathmr{and} {v}_{0}$ are initial position and initial velocity respectively.
From equation (2)
$v \left(t\right) = {\int}_{0}^{t} \left(36 - {t}^{2}\right) \cdot \mathrm{dt}$, on integration yields

$v \left(t\right) = {\left[36 t - {t}^{3} / 3\right]}_{0}^{t} + C$, $C$ being constant of integration.
We obtain $v \left(t\right) = \left(36 t - {t}^{3} / 3 + C\right) - \left(36 \times 0 - {0}^{3} / 3 + C\right)$
or $v \left(t\right) = 36 t - {t}^{3} / 3$
Inserting this value in equation (3), we obtain
$x \left(t\right) = {x}_{0} + {\int}_{0}^{t} \left(36 t - {t}^{3} / 3\right) \cdot \mathrm{dt}$, on integration over the given range of time $t$ yields

$x \left(5\right) = {x}_{0} + {\int}_{0}^{5} \left(36 t - {t}^{3} / 3\right) \cdot \mathrm{dt}$

$x \left(5\right) = {x}_{0} + {\left[36 {t}^{2} / 2 - \frac{1}{3} \cdot {t}^{4} / 4 + {C}_{1}\right]}_{0}^{5}$, where ${C}_{1}$ is constant of integration
$x \left(5\right) = {x}_{0} + \left\{36 {5}^{2} / 2 - \frac{1}{3} \cdot {5}^{4} / 4 + {C}_{1}\right\} - \left\{36 {0}^{2} / 2 - \frac{1}{3} \cdot {0}^{4} / 4 + {C}_{1}\right\}$
or $x \left(5\right) = {x}_{0} + 450 - 52.08 \dot{3}$
or $x \left(5\right) = {x}_{0} + 397.91 \dot{6}$
Inserting this value in equation (1), we obtain
${v}_{\textrm{a v e r a \ge}} = \frac{\cancel{{x}_{0}} + 397.91 \dot{6} - \cancel{{x}_{0}}}{5}$
${v}_{\textrm{a v e r a \ge}} = 79.58 \dot{3} \textrm{u n i t s}$