Question

A car, starting from rest, has a constant acceleration ${\mathrm{a}}_1$ for a time interval ${\mathrm{t}}_1$ during which it covers a distance ${\mathrm{s}}_1$ . In the next time interval ${\mathrm{t}}_2$, the car has a constant retardation ${\mathrm{a}}_2$ and comes to rest after covering a distance ${\mathrm{s}}_2$ in time ${\mathrm{t}}_2$. Which of the following is correct?

• A. $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{s}}_1}{{\mathrm{s}}_2}=\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}$
• B. $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{s}}_2}{{\mathrm{s}}_1}=\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}$
• C. $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{s}}_2}{{\mathrm{s}}_1}=\frac{{\mathrm{t}}_2}{{\mathrm{t}}_2}$
• D. $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{s}}_2}{{\mathrm{s}}_1}=\frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}$

Answer 1

The correct option is D

$\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{s}}_2}{{\mathrm{s}}_1}=\frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}$

Step 1: Given data

A car, starting from rest, has a constant acceleration ${\mathrm{a}}_1$ for a time interval ${\mathrm{t}}_1$ during which it covers a distance ${\mathrm{s}}_1$ .

In the next time interval ${\mathrm{t}}_2$, the car has constant retardation ${\mathrm{a}}_2$ and comes to rest after covering a distance ${\mathrm{s}}_2$ in time ${\mathrm{t}}_2$.

Step 2: Consider case 1

A car, starting from rest, has a constant acceleration ${\mathrm{a}}_1$ for a time interval ${\mathrm{t}}_1$ during which it covers ${\mathrm{s}}_1$ distance. So,

Initial velocity, $\mathrm{u}=0$

The second equation of motion is,

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Here, $\mathrm{s}$ is the distance travelled time $\mathrm{t}$, $\mathrm{u}$ is the initial velocity, and $\mathrm{a}$ is the acceleration of the body in motion.

${\mathrm{s}}_1=\frac{1}{2}{\mathrm{a}}_1{\mathrm{t}}_1^2$ ------- $\left(1\right)$

And velocity after time ${\mathrm{t}}_1,{\mathrm{v}}_1={\mathrm{a}}_1{\mathrm{t}}_1$

Step 3: Consider case 2

In the next interval ${\mathrm{t}}_2$, the car has constant retardation ${\mathrm{a}}_2$ and comes to rest after covering a distance ${\mathrm{s}}_2$ in time interval t₂.

So,

Velocity after time ${\mathrm{t}}_2,{\mathrm{v}}_2={\mathrm{u}}_2-{\mathrm{a}}_2{\mathrm{t}}_2$

Here,

${\mathrm{u}}_2={\mathrm{v}}_1$

$={\mathrm{a}}_1{\mathrm{t}}_1$

${\mathrm{v}}_2={\mathrm{a}}_1{\mathrm{t}}_1-{\mathrm{a}}_2{\mathrm{t}}_2$

But the body comes to rest so ${\mathrm{v}}_2=0$

So,

${\mathrm{a}}_1{\mathrm{t}}_1-{\mathrm{a}}_2{\mathrm{t}}_2=0$

$\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}$ --------- $\left(2\right)$

Now, the distance covered in ${\mathrm{t}}_2$ time is,

${\mathrm{s}}_2=\frac{1}{2}{\mathrm{a}}_2{\mathrm{t}}_2^2$ ------------ $\left(3\right)$

From equations $\left(1\right)$ and $\left(3\right)$, we get,

$\frac{{\mathrm{s}}_1}{{\mathrm{s}}_2}=\frac{{\mathrm{a}}_1{\mathrm{t}}_1^2}{{\mathrm{a}}_2{\mathrm{t}}_2^2}$

From the equation $\left(2\right)$ we get,

$\frac{{\mathrm{s}}_1}{{\mathrm{s}}_2}=\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}$ ------------ $\left(4\right)$

From equations $\left(2\right)$ and $\left(4\right)$ we get,

$\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{s}}_2}{{\mathrm{s}}_1}=\frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}$

Hence, option $\mathrm{D}$ is the correct answer.