Question

a car moving along A straight highway with speed of 72 km per hour is pro to a stop within a distance of 100 M what is the retardation of the car and how long does it take for a car to stop

Answer 1

Inital velocity,u=72km/hr
km into m/s=
$\frac{72 \times 000}{3600}$
=20m/s
Final velocity,v=0m/s(car applies brakes)
Distance=100m
Accelaration=?
Time taken=?

First, we will findout accelaration-

$2as = {v}^{2} - {u}^{2}$
$2 \times a \times 100 = {0}^{2} - {20}^{2}$
$200a = 0 - 400$
$200a = - 400$
$a = \frac{400}{200 }$
$a = - 2$

So the retardation is of = 2ms-2


Lets findout time taken by car to stop-
$a = \frac{v - u}{t}$
$- 2 = \frac{0 - 20}{t}$
$- 2 = \frac{ - 20}{t}$
By cross multiply we get,
$t = \frac{ - 20}{ - 2}$
sign of subtraction will be cancel,
$t = 10$
The time taken by the car to stop is =10 seconds

Answer 2

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u = 72 km/h

v=0

s =100m

v^2-u^2= 2as

-u^2= 2×a×100(v=0)

- 20^2=200a

-400=200a

a=-2 m/s^2

I hope, this will help you

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