Question

hey guys....
Help me in this problem.
A car moving with speed v on the straight track can be stopped in the distance xby applying brakes. If the same car is moving with speed 2v n brakes provide half the retardation then car will stop after traveling the distance




X/4
X/2
8x
X/8​

Answer 1

see this with steps ans 8x

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 - 2ax

acceleration = v²/2x

Now,

according to question

Initial velocity = 2v

retardation = a/2 => v²/4x

Final velocity = 0

stopping distance = ?

Putting in the formula

$0 = 2v {}^{2} - 2( \frac{v {}^{2} }{4x} ) \times distance \\ \\ 8v {}^{2} = \frac{v {}^{2} }{2x} \times distance$

$distance = 8x$

Hope this will help