Question

A car moving with speed 'v' on a straight road can be stopped with a distance 'd' on applying brakes. If the same car is moving with speed '3v' and brakes provide half retardation, then the car will stop after travelling what distance?

Answer 1

Explanation:

6d since v is directly proportional to d so for 3v distance is 3d but brakes produce half retardation so 6d is the ans

Answer 2

Given

1.initial velocity = v

2.final velocity = 0

3.stopping distance = d

4.acceleration = ?

Using 3rd eq. of motion

v² = u² + 2as

Putting values

0 = v² - 2ad

acceleration = v²/2d

Now,

according to question

Initial velocity = 3v

retardation = a/2 => v²/4d

Final velocity = 0

stopping distance = ?

Putting in the formula

$0 = 3v {}^{2} - 2( \frac{v {}^{2} }{4d} ) \times distance \\ \\ 9v {}^{2} = \frac{v {}^{2} }{2d} \times distance$

$distance = 18d$

Hope this will help