A motorcyclist who is moving along an x axis directed toward the east has an acceleration given by $\mathit{a}\mathbf{=}\left(6.1‐1.2t\right)\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$for$\mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{\le }\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$. At$\mathit{t}\mathbf{=}\mathbf{0}$, the velocity and position of the cyclist are$\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ and$\mathbf{7}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{m}$. (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel between$\mathit{t}\mathbf{=}\mathbf{0}$and$\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$?

$$\begin{gathered} a=\left(6.1‐1.2t\right)\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{v}=2.7\mathrm{m}/\mathrm{s} \\ x=7.3\mathrm{m} \end{gathered}$$

If we are given with the displacement equation with respect to time and we differentiate the equation with respect to time, resulted equation will be velocity equation. If we differentiate velocity equation with respect to time, resulted equation will be acceleration equation. If we are given with the acceleration equation with respect to time and we integrate with respect to time, resulted equation will be velocity equation. If we integrate velocity equation with respect to time, resulted equation will be displacement equation.

Formula:

The velocity in general is given by,

$v={\int }_{}^{}adt$

The distance is given by

$x={\int }_{}^{}vdt$

The velocity of the cyclist can be computed as

$$\begin{gathered} v={\int }_{}^{}adt \\ v=\int \left(6.1-1.2t\right) \\ v=6.1t-1.2\left(\frac{t^2}{2}\right)+c \\ v=6.1t-0.6t^{2}c \\ \mathrm{At}t=0\sec ,v=2.7m/sso, \\ 2.7=0-0+c \\ c=2.7 \end{gathered}$$

So, $v=6.1t-0.6t^2+2.7$

The cyclist will have maximum speed when $\frac{dv}{dt}=a=0$

So, by putting $a=0$we will get

$$\begin{gathered} 0=6.1-1.2t \\ t=5.1\sec \\ v=18\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the maximum velocity achieved by the cyclist would be$18\mathrm{m}/\mathrm{s}$

The distance traveled by the cyclist can be calculated as

$$\begin{gathered} x={\int }_{}^{}vdt \\ x=\int 6.1t-0.6t^2+2.7 \\ x=6.1\frac{t^2}{2}-\frac{{\left(0.6\right)}^3}{3}+2.7t+k \\ x=3.05t^2-\left(0.2\right)t^2+2.7t+k \\ \mathrm{At}t=0\sec ,x=7.3m \\ 7.3=0+k \\ k=7.3 \end{gathered}$$

So, the distance traveled by the cyclist between $t=0$and $6.0\sec$.is

$$\begin{gathered} x={\left[3.05t^2-\left(0.2\right)t^2+2.7t+7.3\right]}_0^6 \\ =82.8 \\ ~83\mathrm{m} \end{gathered}$$

Hence, the distance traveled by the cyclist would be 83 m