An electric vehicle starts from rest and accelerates at a rate of$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$in a straight line until it reaches a speed of 20 m/s. The vehicle then slows at a constant rate of$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?

Acceleration during the speeding up is, ${\mathrm{a}}_1=-2.0\mathrm{m}/{\mathrm{s}}^2$

Final velocity during the speeding up is, $\mathrm{v}=20\mathrm{m}/\mathrm{s}$

Acceleration during the speeding down is, ${\mathrm{a}}_2=-1.0\mathrm{m}/{\mathrm{s}}^2$

Kinematic equations can be used to find the distance covered by the vehicle during the acceleration and deceleration of the vehicle. Using the kinematic equation find the time required for the vehicle to stop and also the distance it has traveled.

The equations for the motion of a particle with constant acceleration are,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{i}}+\mathrm{at}\left(\mathrm{i}\right) \\ {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_{\mathrm{i}}^2+2\mathrm{ax}\left(\mathrm{ii}\right) \end{gathered}$$

Here,${\mathrm{v}}_{\mathrm{f}}$ is the final velocity,${\mathrm{v}}_{\mathrm{i}}$ is the initial velocity, a is the acceleration, t is the time and x is the distance.

Here, consider two motions, one from 0 m/s to 20 m/s with acceleration of$2.0\mathrm{m}/{\mathrm{s}}^2$and another from 20 m/s to 0 m/s to with acceleration of$-1.0\mathrm{m}/{\mathrm{s}}^2$

Now, consider the first motion from 0 m/s to 20 m/s with acceleration of$2.0\mathrm{m}/{\mathrm{s}}^2$

Now consider second motion from 20 m/s to 0 m/s with acceleration of $-1.0\mathrm{m}/{\mathrm{s}}^2$.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{i}}+{\mathrm{a}}_2{\mathrm{t}}_2 \\ 0=20+(-1)\times {\mathrm{t}}_2 \\ {\mathrm{t}}_2=20\sec \end{gathered}$$

So total time elapsed from start to stop is,

$$\begin{gathered} \mathrm{t}={\mathrm{t}}_1+{\mathrm{t}}_2\hspace{0.33em}\hspace{0.33em} \\ =10\sec \hspace{0.33em}+\hspace{0.33em}20\sec \hspace{0.33em} \\ =30\hspace{0.33em}\sec \end{gathered}$$

Therefore, the total time that elapses from start to stop is 30 sec .

For first motion.

${{\mathrm{v}}_{\mathrm{f}}}^2={{\mathrm{v}}_{\mathrm{i}}}^2+2{\mathrm{a}}_1{\mathrm{x}}_1$

Solve the equation for distance.

${\mathrm{x}}_1=\frac{{{\mathrm{v}}_{\mathrm{f}}}^2-{{\mathrm{v}}_{\mathrm{i}}}^2}{2{\mathrm{a}}_1}$

Substitute the values of initial and final velocities.

$$\begin{gathered} {\mathrm{x}}_1=\frac{{20}^2-0^2}{2\times 2}\hspace{0.33em}\hspace{0.33em} \\ \hspace{0.33em}\hspace{0.33em}=100\hspace{0.33em}\mathrm{m} \end{gathered}$$

For second motion,

$$\begin{gathered} {\mathrm{x}}_2=\frac{{\mathrm{v}}_{\mathrm{f}}^2-{\mathrm{v}}_{\mathrm{i}}^2}{2{\mathrm{a}}_2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \\ =\frac{0^2-{20}^2}{2\times (-1)} \\ \hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}200\hspace{0.33em}\mathrm{m} \end{gathered}$$

So total distance is,

$$\begin{gathered} \mathrm{x}={\mathrm{x}}_1+{\mathrm{x}}_2\hspace{0.33em} \\ \hspace{0.33em}=100\hspace{0.33em}\mathrm{m}+200\hspace{0.33em}\mathrm{m}\hspace{0.33em}\hspace{0.33em} \\ \hspace{0.33em}=300\mathrm{m} \end{gathered}$$

Thus, the distance travelled by the vehicle from start to stop is 300 m.