Question

A truck on a straight road starts from rest, accelerating at $2m{s}^{-2}$ , until it reaches a speed of $20m{s}^{-1}$ . Then the truck travels at a constant speed for $20s$ until the brakes are applied causing the truck to decelerate uniformly to stop it in additional $5s$ .

(a) For how much time the truck is in motion?

(b) What is the average speed of the truck for its entire motion?

Answer 1

Step 1: Given Data

A truck is moving on a straight road having acceleration, $a=2m{s}^{-2}$

The initial velocity of the truck, $u=0m{s}^{-1}$

The final velocity of the truck, $v=20m{s}^{-1}$

The truck travels at a constant speed for a time, $t_2=20s$

The time required to stop after brakes are applied, $t_3=5s$

Step 2: Determine the total time for which the truck is in motion

Let the time it takes to attain $20m{s}^{-1}$ as the final velocity be $t_1$ .

As we know that the first equation of motion is $\mathit{v}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$

$\Rightarrow t_1=\frac{v-u}{a}$

$\Rightarrow t_1=\frac{20-0}{2}$

$\Rightarrow t_1=10s$

Thus, the total time for which the truck is in motion is $t=t_1+t_2+t_3=10+20+5=35s$

Step 3: Determine the distance covered by the truck for the first $\mathbf{10}\mathbf{}\mathit{s}$

As we know that the second equation of motion is $\mathit{S}\mathbf{=}\mathit{u}\mathit{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$

Let the distance traveled for the first $10s$ be $S_1$

$\Rightarrow S_1=0+\frac{1}{2}\times 2\times {\left(10\right)}^2$

$\Rightarrow S_1=100m$

Step 4: Determine the distance covered by the truck for the next $\mathbf{20}\mathbf{}\mathit{s}$

For next $20s$ the truck moved at a constant speed. Hence, $a=0m{s}^{-1}$

Let the distance traveled for the next $20s$ be $S_2$

On applying the given data in $\mathit{S}\mathbf{=}\mathit{u}\mathit{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$ we get,

$\Rightarrow S_2=20\times 20+0$

$\Rightarrow S_2=400m$

Step 5: Determine the distance covered by the truck for the last $\mathbf{5}\mathbf{}\mathit{s}$

For the last $5s$ the truck gets stopped after applying brakes

Let the distance traveled for the last $5s$ be $S_3$

On applying the equation, $\mathit{S}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(v+u\right)\mathit{t}$

$\Rightarrow S_3=\frac{1}{2}\left(20+0\right)\times 5$

$\Rightarrow S_3=50m$

Thus, the total distance covered by the truck is $S=S_1+S_2+S_3=100+400+50=550m$

Step 6: Determine the average speed of the truck

$\mathit{A}\mathit{v}\mathit{e}\mathit{r}\mathit{a}\mathit{g}\mathit{e}\mathbf{}\mathit{s}\mathit{p}\mathit{e}\mathit{e}\mathit{d}\mathbf{=}\mathbf{}\frac{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{d}\mathbf{i}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}}{\mathbf{T}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{k}\mathbf{e}\mathbf{n}}$

$$\begin{gathered} =\frac{S}{t} \\ =\frac{550}{35} \\ =15.7m{s}^{-1} \end{gathered}$$

Final Answer:

(a) The total time for which the truck is in motion is $\mathbf{35}\mathbf{}\mathit{s}$.

(b) The average speed of the truck for its entire motion is $\mathbf{15}\mathbf{.}\mathbf{7}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.