Question

A body moving from its initial position of rest along a straight line covers $1m$ in $1s$. If it covers $8m$ in the next $2s$, then the body is moving with

• A. uniform velocity
• B. an acceleration of $2m{s}^{-2}$ in the first second
• C. average $t=1s$ to $3s$ of $2m{s}^{-2}$ from acceleration
• D. both (b) and (c)

Answer 1

The correct option is D

both (b) and (c)

Step 1: Given data

As the body starts from rest, therefore, the initial velocity is $u=0m{s}^{-1}$.

Distance covered in $1s$, $s_1=1m$.

Distance covered in next $2s$, $s_2=8m$.

Step 2: Find the acceleration of the body at $\mathbf{1}\mathbf{}\mathit{s}\mathbf{}$

Assume the acceleration of the body in $1s$as $a_{1}m{s}^{-2}$.

Since we know the second equation of motion is $\mathit{s}\mathbf{=}\mathit{u}\mathit{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$,

So for the case when the body covers $1m$ in $1s$, the acceleration is:

$\Rightarrow 1=0+\frac{1}{2}{a}_1{\left(1\right)}^2$

$\Rightarrow 1=\frac{1}{2}{a}_1{\left(1\right)}^2$

$\Rightarrow a_1=2m{s}^{-2}$

As the body is accelerating, the body is not moving with a uniform velocity.

Step 3: Find the velocity of the body at $\mathbf{1}\mathbf{}\mathit{s}$.

Assume the velocity of the body in $1s$as $v_{1}m{s}^{-1}$.

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

For finding the velocity of the body in $1s$we have to put all the required values in the above equation.

$\Rightarrow v=0+2\left(1\right)$

$\Rightarrow v_1=2m{s}^{-1}$

Step 4: Find the velocity and acceleration of the body for next $\mathbf{2}\mathbf{}\mathit{s}$.

Assume the velocity of the body for the next $2s$as $v_{2}m{s}^{-1}$and the acceleration of the body for the next $2s$as $a_{2}m{s}^{-2}$.

Also, we know the second equation of motion is $\mathit{s}\mathbf{=}\mathit{u}\mathit{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$.

$\Rightarrow 8=2\left(2\right)+\frac{1}{2}{a}_2{\left(2\right)}^2$

$$\begin{gathered} \Rightarrow 8=4+2a_2 \\ \Rightarrow 8-4=2a_2 \\ \Rightarrow 2a_2=4 \end{gathered}$$

$\Rightarrow a_2=2m{s}^{-2}$

Therefore, the velocity of the body for the next $2s$ (at $3s$)by using the first equation of motion is,

$$\begin{gathered} \Rightarrow v_2=2+2\left(2\right) \\ \Rightarrow v_2=2+4 \end{gathered}$$

$\Rightarrow v_2=6m{s}^{-1}$

Step 5: Find the average acceleration from $\mathbf{1}\mathbf{}\mathit{s}$ to $\mathbf{3}\mathbf{}\mathit{s}$ time interval.

The formula for average acceleration is $\mathit{a}\mathit{v}\mathit{e}\mathit{r}\mathit{a}\mathit{g}\mathit{e}\mathit{}\mathit{a}\mathit{c}\mathit{c}\mathit{e}\mathit{l}\mathit{e}\mathit{r}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{}\mathit{=}\mathit{}\frac{\mathit{c}\mathit{h}\mathit{a}\mathit{n}\mathit{g}\mathit{e}\mathit{}\mathit{i}\mathit{n}\mathit{}\mathit{v}\mathit{e}\mathit{l}\mathit{o}\mathit{c}\mathit{i}\mathit{t}\mathit{y}}{\mathit{t}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathit{}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathit{}\mathit{t}\mathit{a}\mathit{k}\mathit{e}\mathit{n}}$.

$$\begin{gathered} \Rightarrow averageacceleration=\frac{v_2-v_1}{3-1} \\ =\frac{6-2}{3-1} \\ =\frac{4}{2} \\ =2m{s}^{-2} \end{gathered}$$

Therefore the average acceleration from $t=1s$ to $3s$ is $2m{s}^{-2}$.

Hence, option D is the correct answer.