# What is the average speed of an object that is moving at 5 m/s at t=0 and accelerates at a rate of a(t) =t-4 on t in [4,5]?

May 11, 2016

${v}_{a} = \frac{17}{6} \text{ } \frac{m}{s}$

#### Explanation:

$\frac{d v}{d t} = a \left(t\right)$

$d v = a \left(t\right) \cdot d t$

$\int d v = \int a \left(t\right) \cdot d t$

$v = \int \left(t - 4\right) \cdot d t$

$v = {t}^{2} / 2 - 4 t + C \text{ "t=0 ;v=5 m/s" ; } c = 5 \frac{m}{s}$

$v = {t}^{2} / 2 - 4 t + 5$

$\frac{d x}{d t} = v \left(t\right)$

$d x = v \left(t\right) \cdot d t$

$\int d x = {\int}_{4}^{5} v \left(t\right) \cdot d t$

$x = {\int}_{4}^{5} \left({t}^{2} / 2 - 4 t + 5\right) d t$

$x = | \frac{1}{2} \cdot {t}^{3} / 3 - 4 \cdot {t}^{2} / 2 + 5 \cdot t {|}_{4}^{5}$

$x = \left(\frac{1}{2} \cdot {5}^{3} / 3 - 2 \cdot {5}^{2} + 5 \cdot 5\right) - \left(\frac{1}{2} \cdot {4}^{3} / 3 - 2 \cdot {4}^{2} + 5 \cdot 4\right)$

x=125/6-50+25)-(64/6-32+20)

$x = \frac{125}{6} - 25 - \frac{64}{6} + 12 = \frac{61}{6} - 13$

$x = \frac{61 - 78}{6}$

$x = | - \frac{17}{6} |$

$x = \frac{17}{6} m e t e r s$

${v}_{a} = \frac{x}{t}$

${v}_{a} = \frac{\frac{17}{6}}{5 - 4}$

${v}_{a} = \frac{17}{6} \text{ } \frac{m}{s}$