Question

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

Answer 1

Answer:

See the attachment

Explanation:

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