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Question

A particle starts moving along a line from rest and comes to rest after moving distance d. During its motion, it had a constant acceleration f over 2/3 of the distance and covered the rest of the distance with constant retardation. The time taken to cover the distance is:

A
2d3f
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B
2d3f
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C
3df
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D
3d2f
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Solution

The correct option is A 3df
Let velocity at B be v.
From A to B :
Initial velocity, u=0
Acceleration, a=f
Time taken, t=t1
Using Newton's first equation of motion,
v=u+at
Velocity at B, v1=0+ft1=ft1
v1=ft1 ....(1)
Using Newton's second equation of motion,
s=ut+12at2
Distance covered, s=2d3
2d3=0+12ft21
t1=4d3f
t1=2d3f .....(b)

From B to C :
Initial velocity, u=v1
Final velocity, v=0
Acceleration, a
Time taken, t=t2
Using Newton's first equation of motion,
v=u+at
0=v1+at2
=v=at2 ....(2)
From (1) and (2), we get
a=ft1t2
Using Newton's third equation of motion
2as=v2u2
2ad3=02v21
d3=v22a=(ft1)22×(ft1/t2)=ft1t22 ....(a)
From (a) and (b), d3=ft22×4d3f
t2=d3f
Thus total time taken, t=t1+t2=2d3f+d3f=3d3f
t=3df

2085264_738276_ans_bbdc7c25dd544fbc9c699fbef528e70f.png

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