Question

A particle starts from rest and moves with a constant acceleration of $a_{1}m{s}^{-2}$ for sometime and then decelerates to come to rest at constant acceleration of $a_{2}m{s}^{-2}$. If the time for which particle is in motion is $t$, then the total distance travelled by the particle is

• A. $\frac{a_1{a}_2{t}^2}{a_1+a_2}$
• B. $\frac{a_1{a}_2{t}^2}{2(a_1+a_2)}$
• C. zero
• D. $\frac{(a_1+a_2)t^2}{2}$

Answer 1

The correct option is B

$\frac{a_1{a}_2{t}^2}{2(a_1+a_2)}$

Step 1: Given

Acceleration$=a_{1}m{s}^{-2}$

Deceleration$=a_{2}m{s}^{-2}$

Total time taken$=ts$

Step 2: Find the velocity

Let the time taken for acceleration be $t_{1}s$ and deceleration be $t_{2}s$

As we know, $\mathit{a}\mathit{c}\mathit{c}\mathit{e}\mathit{l}\mathit{e}\mathit{r}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{=}\frac{\mathbf{v}\mathbf{e}\mathbf{l}\mathbf{o}\mathbf{c}\mathbf{i}\mathbf{t}\mathbf{y}}{\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}}$

∴ During acceleration: $a_1=\frac{v}{t_1}\Rightarrow t_1=\frac{v}{a_1}s$

∴ During deceleration: $a_2=\frac{v_2}{t_2}\Rightarrow t_2=\frac{v}{a_2}s$

Now, the total time $ts$ is:

$$\begin{gathered} \Rightarrow t=t_1+t_2 \\ t=\frac{v}{a_1}+\frac{v}{a_2} \\ t=v\frac{a_1+a_2}{a_1{a}_2} \\ v=\frac{a_1{a}_{2}t}{a_1+a_2}m{s}^{-1} \end{gathered}$$

Step 3: Calculate the distance

As we know, $\mathit{v}\mathit{e}\mathit{l}\mathit{o}\mathit{c}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{d}\mathbf{i}\mathbf{s}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}}{\mathbf{t}\mathbf{i}\mathbf{m}\mathbf{e}}$

$$\begin{gathered} \therefore v=\frac{s}{t} \\ \Rightarrow s=vt \\ s=\frac{a_1{a}_{2}t}{a_1+a_2}\times t \\ s=\frac{a_1{a}_2{t}^2}{a_1+a_2}m \end{gathered}$$

The total distance traveled by the particle is $\frac{{\mathbf{a}}_{\mathbf{1}}{\mathbf{a}}_{\mathbf{2}}\mathbf{}\mathbf{}{\mathbf{t}}^{\mathbf{2}}}{{\mathbf{a}}_{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{2}}}\mathbf{}\mathit{m}$. Hence, option B is the correct answer.