Question

The speed of the car is $v$, the minimum distance over which it can be stopped is $x$. If the speed becomes $nv$, then the minimum distance in which it can be stopped in the same time is

• A. $x/n$
• B. $nx$
• C. $x/n^2$
• D. $n^{2}x$

Answer 1

The correct option is B

$nx$

Step 1: Given values,

Speed of a car is $v$.

The minimum distance over which we can stop is $x.$

Step 2: Calculate the minimum distance,

We know that a change in velocity with respect to time is called acceleration.

Acceleration, $a=\frac{v-u}{t}$

In the first case, during retardations,
$a=\frac{0-v}{t}=-\frac{v}{t}$

From the second equation of motion,

$x=vt+\frac{1}{2}a{t}^2=vt-\frac{vt}{2}=\frac{vt}{2}$
In the second case, during retardation,
$a^{\text{'}}=\frac{0-nv}{t}=-\frac{nv}{t}$

From the second equation of motion,

$\therefore s=nvt-\frac{1}{2}\frac{nv}{t}{t}^2=\frac{nvt}{2}$
But

$$\begin{gathered} \frac{vt}{2}=x \\ \therefore s=nx \end{gathered}$$

Thus, the minimum distance in which it can be stopped at the same time is $nx.$

Hence, option B is the correct answer.