Question

Brakes applied to a car produce a uniform retardation of 90 cm/s^2. If the car was travelling with a velocity of 27 m/s, then the distance it will cover before coming to rest was

503 m

405 m

610 m

none

Clear selection


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Answer 1

Given The Retardation offered to the Car is

90cm/s^2 or

$90cms {}^{ - 2} \: or \: 0.9ms {}^{ - 2}$

Also that the cae

r is travelling at a velocity of 27m/s

The distance car travells befire coming to rest is given by

Third equation of motion

$v {}^{2} - u {}^{2} = 2as$

Where v is final velocity which is Zero

v is final velocity which is Zero as car comes to rest after few seconds of applications of brakes

U is initial velocity with which car was travelling

a is the negative of retardation offered to the car

S is the distance covered

$0 {}^{2} - (27) {}^{2} = 2 \times( - 0.9) \times s$

$- 729 = - 1.8s$

$729 = 1.8s$

$s = \frac{729}{1.8}$

$\huge{\pink{S=405m}}$

Answer 2

Explanation:

use 3rd equation of motion