# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) = 5t+1 from t in [0, 6]?

May 31, 2017

$33 \text{m"/"s}$

#### Explanation:

To find the average speed of the object, we have to know both the total distance traveled ($\Delta x$) and the time ($6 \text{s}$, according to the given interval).

We need to integrate this acceleration equation in order to find the position equation. The integral of ${t}^{n}$ ($n$ being $1$ in this case, the exponent of $t$ in the "$5 t$" quantity) is denoted $\int {t}^{n} \mathrm{dt}$, and is equal to $\frac{1}{n + 1} {t}^{n + 1}$, which in this case is

$\frac{1}{1 + 1} {t}^{1 + 1} = \frac{1}{2} {t}^{2}$

Doing the same for the second quantity ($1$), we have

$\frac{1}{0 + 1} {t}^{0 + 1} = 1 {t}^{1} = t$

Therefore, the first integral of the acceleration equation, which is the velocity equation, is

${v}_{x} \left(t\right) = {v}_{0 x} + \frac{1}{2} \left(5\right) {t}^{2} + t = \frac{5}{2} {t}^{2} + t$

because the initial velocity and position are assumed to be $0$.

Since this is the velocity equation, we need to integrate this again to find the position equation. Following the same procedure as before, our result is

$x \left(t\right) = {x}_{0} + \frac{1}{3} \left(\frac{5}{2}\right) {t}^{3} + \frac{1}{2} {t}^{2} = \frac{5}{6} {t}^{3} + \frac{1}{2} {t}^{2}$

Now that we have our position equation, let's find the position of the particle at the time $t = 6 \text{s}$:

x(6) = 5/6(6)^3+ 1/2(6)^2 = color(red)(198"m"

Since the velocity and position are always increasing here, the average speed is the same as the average velocity, which is given by the equation

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

Since our initial position and time are both $0$, the average speed of the object over the interval $t \in \left[0 , 6\right]$ is

v_(av-x) = (color(red)(198"m"))/(6"s") = color(blue)(33"m"/"s"