Question

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

Answer 1

1st speed = vStopped in distance = d2nd speed =3vRetardation =1/2What is distance covered by car while braking or stopping?By puting 3rd law of motion i.e v^2 = u^2+2asSo a=v^2÷2da=1÷2(v^2÷2d)=v^2÷4dd=(3v)^2÷2d=9v^2÷2v^2÷4d=9×4d÷2= 18d
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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 = 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