Question

A car accelerates from rest at $2m{s}^{-2}$ for $5s$ , travels at a constant speed of for $7s$, accelerates at $1m{s}^{-2}$ for $10s$ and then decelerates to rest at $3m{s}^{-2}$ . Determine the average speed of car for entire journey.

Answer 1

Step 1: Given Data

The acceleration of the car in the first $5s$ is $a_1=2m{s}^{-2}$.

The acceleration of the car for the next $10s$ is $a_3=1m{s}^{-2}$ .

The deceleration of the car at last is $a_4=3m{s}^{-2}$.

Step 2: Find the speed of the car for the first $\mathbf{5}\mathbf{}\mathit{s}$ .

As we know, the second equation of motion is $\mathit{S}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$. On putting the values we get:

$s_1=0+\frac{1}{2}\left(2\right){\left(5\right)}^2$

$s_1=25m$

As we know, the first equation of motion $\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$. On putting the values we get:

$v_1=2\times 5=10m{s}^{-1}$

Step 3: Find the speed of the car for the next $\mathbf{7}\mathbf{}\mathit{s}$.

Let us assume the acceleration of the car for next $7s$ as $a_2$.

The car is moving at a constant speed. Hence, $a_2=0m{s}^{-2}$.

From the second equation of motion $\mathit{S}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$, we get:

$s_2=v_{1}t+\frac{1}{2}{a}_2{t}^2$

$s_2=10\times 7+0$

$s_2=70m$

Step 4: Find the speed of the car for the next $\mathbf{10}\mathbf{}\mathit{s}$.

From the second equation of motion $\mathit{S}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$ we get:

$s_3=v_{1}t+\frac{1}{2}{a}_3{t}^2$

$s_3=10\times 10+\frac{1}{2}\times 1\times {\left(10\right)}^2$

$s_3=150m$

As the first equation of motion is $\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$ then the final velocity of the car is:

$v_3=10+1\left(10\right)$

$v_3=20m{s}^{-1}$

Step 5: Find the speed of the car for the remaining seconds

The car gets decelerated by $a_4=-3m{s}^{-2}$

Assume the time taken by the car after $10s$ to stop as $t_4$.

From the first equation of motion $v=u+at$ we get $t=\frac{v-u}{a}$.

$$\begin{gathered} \Rightarrow t_4=\frac{0-20}{-3} \\ \Rightarrow t_4=\frac{20}{3}s \end{gathered}$$

From the second equation of motion, we get:

$s_4=20\times \frac{20}{3}+\frac{1}{2}\left(-3\right){\left(\frac{20}{3}\right)}^2$

$s_4=\frac{200}{3}m$

Step 6: Find the total distance covered by the car.

Thus, the total distance covered by the car is $s=s_1+s_2+s_3+s_4$

$\Rightarrow s=25+70+150+\frac{200}{3}$

$\Rightarrow s=\frac{935}{3}m$

Step 7: Find the total time taken by the car.

The total time required to cover the journey is $t=5+7+10+t_4$.

$\Rightarrow t=22+\frac{20}{3}$

$\Rightarrow t=\frac{86}{3}s$

Step 8: Determine the average speed of the car for the entire journey

The average speed of the car $=\frac{totaldis\tan ce}{totaltimetaken}$.

$$\begin{gathered} =\frac{s}{t} \\ =\frac{935}{86} \\ =10.87m{s}^{-1} \end{gathered}$$

Final Answer:

The average speed of a car for the entire journey is $\mathbf{10}\mathbf{.}\mathbf{87}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.