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in Laws of Motion and Friction by (15 points)
9. A body moving with initial velocity \( 83 m / s \) moved under constant retardation of \( 3 m / s ^{2} \). Find its velocity after \( 2 sec \). Time after which it will stop and the aíctance it will travel before coming to rest.

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2 Answers

+2 votes
by (1.5k points)

As per question,

u = 83m/s

a = -3m/s2

t = 2 s

v = ?

v = u + at

v = 83 + (-3) x 2

v = 83 - 6 = 77m/s     Ans.....

To find the time after which it stops i.e. v becomes 0.

again v = u + at

0 = 83 - 3 x t

t = 83/3 = 27.66s   Ans....

To find the distance travelled before coming to rest,

v= u+ 2ax   ( also x = ut + 1/2 atcan be used)

02 = 83+ 2(-3) x

x = - 832/-6 = 1148.16m    Ans......

+1 vote
by (18.1k points)

Given initial velocity,

u = 83 m/s

a = 3 m/s2

t = 2 sec

v = u + at

v = 83 + 3 x 2

after 2 sec velocity v = 89 m/s.

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