Question

A body moves with a velocity 30m/s applies breaks and stop in 20 sec. Find deacceleratiom and displacement covered. ​

Answer 1

Hi there,

:)

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Retardation => acceleration in the direction opposite to velocity, continually   reducing the length of the velocity vector until its direction becomes the same as the acceleration(reverses), (i.e; previously retardation).....

displacement in 20s

S = ut -1/2 at²

S =30 x 20 - 1/2 a x 20²

S= 600 -a/2 x 400

S = 100 (6-2a)

now.... here we can use the work energy theorem,

W= FS cos x

(F = force , S= displacement, x= angle of force wrt displacement)

and according to work energy theorem,

W(initial)=W(final)

1/2 m v² = work (in the simple form)

1/2 m v² = F S cos x

F= retarding forc(=> acceleration is opposite to velocity) = m a

a= (here is retardation)

cos x = -1 (x=180 degree) because... (force acts opposite to displacement)

1/2 m v² = m a cos x

1/2 x30² = a x 100 (6-2a) x -1

from this we get quadratic...

4a²-12a-9=0

which has perfect root..... (b²-4ac)=0

so... solving....

a = 3/2 = 1.5 m/s

substituting in [S = 100 (6-2a) ]

we get

S = 100 (6 - 3)

S = 300m....

Cheers:)

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