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 $t$, find the time for which the particle accelerates with acceleration $a_1$

• A. $\frac{a_2}{a_1+a_2}\times t$
• B. $\frac{a_1}{a_2}\times t$
• C. $\frac{a_1}{a_1+a_2}\times t$
• D. $\frac{a_2}{a_1}\times t$

Answer 1

The correct option is A

$\frac{a_2}{a_1+a_2}\times t$

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 2: Calculate the time

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

$$\begin{gathered} \Rightarrow t_1=\frac{v}{a_1} \\ =\frac{a_1{a}_{2}t}{a_1+a_2}\times \frac{1}{a_1} \\ =\left(\frac{a_{2}t}{a_1+a_2}\right)s \end{gathered}$$

The time for which the particle accelerates with acceleration ${\mathit{a}}_{\mathbf{1}}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{2}}$ is $\left(\frac{a_{2}t}{a_1+a_2}\right)\mathbf{}\mathit{s}$ Hence, option A is the correct option.