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Question

A particle has an initial velocity of 9m/s due east and a constant acceleration of 2m/s2 due west. The distance covered by the particle in the fifth second of its motion is:

A
0
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B
0.5m
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C
2m
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D
0.25m
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Solution

The correct option is B 0.5m

Here the particle velocity will become zero in 4.5 seconds, as

v=u+at0=92×tt=4.5sec

Displacement in the same period will be

v2u2=2as81=2ss4.5=20.25m

Let us call the point of velocity being zero as Point A.

now body will start moving backwards,

displacement in next 0.5 sec from point A,

s=ut+12at2s0.5=12(2)(0.25)2=0.25m

hence distance covered is 0.25 m

(-ve sign indicates body moved in backward direction. It is to be noted that displacement and distance are not to be confused to be same)

Distance traveled in 4 seconds

s4=9×4+12(2)(4)2s4=3616=20

Thus,

distance covered in 5th second is

D=(s4.5+|s0.5|s4)=20.5+0.2520=0.5


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