Question

A car moving at 72km/h is stopped in 5 seconds.If it retards uniformly find the distance travelled by the car before it stops.Also find its retardation.

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Answer 1

Answer:

Given

initial velocity = 72km/h

= 72*5/18 m/s

= 20 m/s

final velocity = 0

time taken = 5 s

from the motion of equation

v = u + at

0 = 20 + a*5

a = -20 / 5

a = -4m/s

retardation = 4m/s

from the second law of mation

v^2 = u^2 + 2aS

0^2 = 20^2 - 2*4*S

0 = 40 - 8S

8S = 40

S = 40 / 8

S = 5 m

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Answer 2

Answer:

given,

initial velocity(u)=72 kmph = 72×5/18= 20m/s

final velocity(v)=0m/s

time(t)=5 s

from the equation of motion,

v=u+at

0=20+a(5)

-20=5a

a=-20/4

a=-5

now from the equation of motion,

v^2-u^2=2as

0^2-(20)^2=2(-5)(s)

-400=-10s

s=-400/-10

s=40m

there fore,

retardation=-5 m/s^2

distance covered before being stopped = 40m