IIT JEE , AIEEE Forum - Mathematics , Vectors

ramanarkar

A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r. The value of t is given by

a) 2s divided by 1/f + 1/r

b) square root of 2s(f+r)

c) square root of 2s(1/f + 1/r)

d) 2s(1/f + 1/r)

saipragadeesh

i hope so the ans is 'a'. but not exactly.

amar.gupta

Dear,

Let in the first part of journey : distance traveled = s₁

and time taken =t₁and speed at the end of first part =v₁

in the second part of journey : distance traveled = s₂

and time taken =t₂

t=t₁+t₂

and s=s₁+s₂

by simple velocity eq. v=u+at and s=ut+(1/2)at²

for first part:

v₁ = 0 + f t₁ .....[1]

and s₁ = 0 + (1/2) f (t₁)² ........[2]

similarly for second part:

0 = v₁- r t₂

or v₁ =r t₂.....[3]

and s₂ = v₁ t₂ - ( 1/2) r (t₂)² ...[4]

from [3] : s₂ = r t₂ t₂ - ( 1/2) r (t₂)²

or s₂= ( 1/2) r (t₂)² ..........[5]

from [1] and [3] :

t₁ / t₂ = r / f ...........[6]

from [5] and [6]

s₂ = (1/2) (f²/ r) (t₁)²

as s₁+s₂ = s = (1/2) (f²/ r) (t₁)²+(1/2) f (t₁)²s = f (1/2)(t₁)² [ (f / r) + 1]

s = (t₁)² (f + r)f / 2 r

or t₁ = sq rt [ 2s r / f (f + r) ]

now t=t₁+t₂or t= t₁+ ( f / r ) t₁

or t= [(r + f ) / r ] t₁

or t= [(r + f ) / r ] sq rt [ 2s r / f (f + r) ]

or t= sq rt [ 2s (f + r) / f r ]

ANS C

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