# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) =t/6 from t in [0, 1]?

Jan 4, 2016

You also need the initial speed of the object ${u}_{0}$. The the answer is:

${u}_{a v} = 0.042 + {u}_{0}$

#### Explanation:

Definition of acceleration:

$a \left(t\right) = \frac{\mathrm{du}}{\mathrm{dt}}$

$a \left(t\right) \cdot \mathrm{dt} = \mathrm{du}$

${\int}_{0}^{t} a \left(t\right) \mathrm{dt} = {\int}_{{u}_{0}}^{u} \mathrm{du}$

${\int}_{0}^{t} \left(\frac{t}{6}\right) \mathrm{dt} = {\int}_{{u}_{0}}^{u} \mathrm{du}$

$\frac{1}{6} {\int}_{0}^{t} \left(t\right) \mathrm{dt} = {\int}_{{u}_{0}}^{u} \mathrm{du}$

$\frac{1}{6} \left({t}^{2} / 2 - {0}^{2} / 2\right) = u - {u}_{0}$

$u \left(t\right) = {t}^{2} / 12 + {u}_{0}$

To find the average speed:

$u \left(0\right) = {0}^{2} / 12 + {u}_{0} = {u}_{0}$

$u \left(1\right) = {1}^{2} / 12 + {u}_{0} = \frac{1}{12} - {u}_{0}$

${u}_{a v} = \frac{{u}_{0} + {u}_{1}}{2}$

${u}_{a v} = \frac{{u}_{0} + \frac{1}{12} + {u}_{0}}{2}$

${u}_{a v} = \frac{2 {u}_{\oplus} \frac{1}{12}}{2}$

${u}_{a v} = \frac{2 {u}_{0}}{2} + \frac{\frac{1}{12}}{2}$

${u}_{a v} = {u}_{0} + \frac{1}{24}$

${u}_{a v} = 0.042 + {u}_{0}$