# What is the average speed of an object that is moving at 18 m/s at t=0 and accelerates at a rate of a(t) =t-8 on t in [1,4]?

Aug 14, 2017

${v}_{\text{avg"=1.5" m"//"s}}$

#### Explanation:

We are given the following information:

• $\mapsto \left\mid \vec{v} \right\mid = 18 \text{ m"//"s at } t = 0$

• $\mapsto a \left(t\right) = t - 8$

• $\mapsto t \in \left[1 , 4\right]$

We can begin by recognizing that acceleration is the derivative of velocity, or equivalently, velocity is the antiderivative/integral of acceleration.

We can take the integral of the provided equation for acceleration to find the equation for velocity at time $t$.

$v \left(t\right) = \int \left(t - 8\right) \mathrm{dt}$

$\implies v \left(t\right) = \frac{1}{2} {t}^{2} - 8 t + C$

We can use that $v = 18$ when $t = 0$ to find the constant $C$:

$18 = \frac{1}{2} {\left(0\right)}^{2} - 8 \left(0\right) + C$

$\implies C = 18$

Therefore, we have: $v \left(t\right) = \frac{1}{2} {v}^{2} - 8 t + 18$

Average velocity is given by:

${v}_{\text{avg}} = \frac{\Delta s}{\Delta t}$

We can take the integral of $v \left(t\right) , t \in \left[1 , 4\right]$ to find the position $\Delta s$.

$\Delta s = {\int}_{1}^{4} \left(\frac{1}{2} {t}^{2} - 8 t + 18\right) \mathrm{dt}$

$\implies = \frac{1}{6} {t}^{3} - 4 {t}^{2} + 18 t {|}_{1}^{4}$

$\implies \frac{1}{6} {\left(4\right)}^{3} - 4 {\left(4\right)}^{2} + 18 \left(4\right) - \frac{1}{6} {\left(1\right)}^{3} + 4 {\left(1\right)}^{2} - 18 \left(1\right)$

$\implies \frac{64}{6} - 64 + 72 - \frac{1}{6} + 4 - 18$

$\implies \frac{9}{2} \text{ m}$

Substituting this into our equation for average velocity:

v_"avg"=(9/2" m")/((4-1)"s")

$\implies = \left(\frac{9}{2} \text{ m")/(3" s}\right)$

$\implies \textcolor{\mathrm{da} r k b l u e}{\frac{3}{2} \text{ m"//"s}}$

$\implies = 1.5 \text{ m"//"s}$

Speed is the magnitude of velocity, and so the answer remains the same.