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 [2, 4]?

Mar 4, 2016

Use the definition of acceleration and know that with respect to time, $u \left(0\right) = 0$ because it is still. Also, you should give units of measurement (e.g. $\frac{m}{s}$). I didn't use any because you didn't give me.

${u}_{a v e r} = 14$

Explanation:

Being still at $t = 0$ means that for $u = f \left(t\right) \to u \left(0\right) = 0$
Starting from the acceleration definition:

$a = \frac{\mathrm{du}}{\mathrm{dt}}$

$t + 3 = \frac{\mathrm{du}}{\mathrm{dt}}$

$\left(t + 3\right) \mathrm{dt} = \mathrm{du}$

${\int}_{0}^{t} \left(t + 3\right) \mathrm{dt} = {\int}_{0}^{u} \mathrm{du}$

${\int}_{0}^{t} t \mathrm{dt} + {\int}_{0}^{t} 3 \mathrm{dt} = {\int}_{0}^{u} \mathrm{du}$

${\left[{t}^{2} / 2\right]}_{0}^{t} + 3 {\left[t\right]}_{0}^{t} = {\left[u\right]}_{0}^{u}$

$\left({t}^{2} / 2 - {0}^{2} / 2\right) + 3 \left(t - 0\right) = u - 0$

$u \left(t\right) = {t}^{2} / 2 + 3 t$

So the average velocity between times 2 and 4 is:

${u}_{a v e r} = \frac{u \left(2\right) + u \left(4\right)}{2}$

$u \left(2\right) = {2}^{2} / 2 + 3 \cdot 2 = 8$

$u \left(4\right) = {4}^{2} / 2 + 3 \cdot 4 = 20$

Finally:

${u}_{a v e r} = \frac{8 + 20}{2}$

${u}_{a v e r} = 14$