Question

A car moving with speed v on a straight track can
be stopped in a distance x by applying brakes. If
same car is moving with speed 2v and brakes
provide half the retardation then car will stop after
travelling distance​

Answer 1

Answer:

8x

Explanation:

v^2 = 2ax

x = v^2 / 2a

v is doubled and a is halved

x' = (2v)^2 / 2(a/2)

= 8 [v^2/ 2a ]

x' = 8x

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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 = 2v

retardation = a/2 => v²/4d

Final velocity = 0

stopping distance = ?

Putting in the formula

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

$distance = 8d$

Hope this will help