Question

A car moving along a straight highway with a speed of 72 km/h is brought to rest with in a distance of 100m. What is the retardation of the car and how long does it takes to stop.

Answer 1

HEY BUDDY..!!!

HERE'S THE ANSWER..

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▶️ First we'll find retardation by

⏺️ ( v )^2 = ( u )^2 + 2 × a × s

✔️ Where ,

⏺️s ( displacement ) = 100 m

⏺️v ( final velocity ) = 0 //coming to rest

⏺️u ( initial velocity ) = 72 Km / h
=> ( 72 × 5 / 18 ) m / s = 20 m / s

⏺️a ( retardation ) = ?

✔️ Putting values in formula

=> ( v )^2 = ( u )^2 + 2 × a × s

=> ( 0 )^2 = ( 20 )^2 + 2 × a × 100

=> - ( 20 )^2 = 2 × a × 100

=> - 400 = a × 200

=> a = - 2 m / s^2 ( it is negative because it is retarding )

✔️ Now we have to find time ( t )

⏺️ We will use [ v = u + a × t ]

=> v = u + a × t

=> 0 = 20 - ( 2 × t )

=> 20 = 2 × t

=> t = 10 seconds

▶️the retardation of the car is ( - 2 m / s^2 ) and take 10 seconds to stop .

HOPE HELPED.

JAI HIND !!!

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Answer 2

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Initial velocity of the car, u = 126 km/h = 35 m/s

Final velocity of the car, v = 0

Distance covered by the car before coming to rest, s = 200 m

Retardation produced in the car = a

From third equation of motion, a can be calculated as:

V^2 - U^2 = 2aS

(0)^2 - (35)^2 = 2 . a . 200

a = ( 35×35)/(2× 200)

a= 3.06 m/ s^2

From first equation of motion, time (t) taken by the car to stop can be obtained as:

V = U + at

t = (V- U )/a = (-35)/(-3.06) = 11.44 sec

I hope, this will help you

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