Question

A car moving at a speed u is stopped in a certain distance when the brakes produce a deceleration a. If the speed of the car is nu, what must be the deceleration of the car to stop it in the same distance?

• A. $\sqrt{n}a$
• B. $na$
• C. $n^{2}a$
• D. $n^{3}a$

Answer 1

Answer: (C)

Step 1, Given data

Speed of the car = u

Retardation = a

The new speed of the car = nu

Step 2, Finding retardation

From the third equation of motion, $v^2=u^2-2as$

Putting all the values

$\Rightarrow 0=u^2-2as$
$\Rightarrow s=\frac{u^2}{2a}$
So, the distance over which the car is stopped is given by $s=\frac{u^2}{2a}$.

Now,

If initial velocity is $nu$, let the value of deceleration to stop the car in the same distance be $a^{\prime }$.

We can write,

$v^2=u^2+2as$

Putting all the values

$0={\left(nu\right)}^2+2\times a^{\prime }\times \frac{u^2}{2a}$

After solving

$a^{\prime }=-a{n}^2$

Hence the new retardation is $a^{\prime }=-a{n}^2$