Speeds of two identical cars are u and 4u at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is :

Sign in

Motion in A Straight Line question

2002 · Shift 0 · Q176

Motion in A Straight Line question

2002 · Shift 0 · Q176

View written solutionFree

Correct answer: D

Use the stopping-distance relation

When a car is brought to rest under the same retardation $a$ , we use

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

For stopping, final velocity $v=0$ .

So,

$0 = u^2 - 2as$

hence,

$s = \frac{u^2}{2a}$

This shows that stopping distance is proportional to the square of the initial speed:

$s \propto u^2$

Apply to the two cars

For the first car, speed $=u$

$s_1 \propto u^2$

For the second car, speed $=4u$

$s_2 \propto (4u)^2 = 16u^2$

Therefore,

$s_1:s_2 = u^2:16u^2 = 1:16$

Match with the options

$1:16$ corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

More from Motion in A Straight Line

Sign in

Motion in A Straight Line question

2002 · Shift 0 · Q176

Motion in A Straight Line question

2002 · Shift 0 · Q176

View written solutionFree

Correct answer: D

Use the stopping-distance relation

When a car is brought to rest under the same retardation $a$ , we use

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

For stopping, final velocity $v=0$ .

So,

$0 = u^2 - 2as$

hence,

$s = \frac{u^2}{2a}$

This shows that stopping distance is proportional to the square of the initial speed:

$s \propto u^2$

Apply to the two cars

For the first car, speed $=u$

$s_1 \propto u^2$

For the second car, speed $=4u$

$s_2 \propto (4u)^2 = 16u^2$

Therefore,

$s_1:s_2 = u^2:16u^2 = 1:16$

Match with the options

$1:16$ corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.