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Question

A man loses 20% of his velocity after running through 108 m. If the rate by which he is losing his velocity remains constant, what is the maximum distance he will cover?

A
218 m
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B
300 m
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C
324 m
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D
192 m
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Solution

The correct option is B 300 m
Let the initial velocity be u. We know the man loses 20% of his velocity after running 108 m.
Hence, velocity at this moment=80% of u=0.8u
Final velocity v=0.8u
Also s=108 m
Using v2=u2+2as,
We get,
(0.8u)2=u2+2a×(108)
216a=0.64u2u2=0.36u2
a=0.36216u2=1600u2

Now let's assume that the man ran a total distance of x m.
We know,
initial velocity =u
final velocity v=0
Distance covered s=x
Acceleration a=1600u2
Using v2=u2+2as
We get,
0=(u)2+2ax
0=u2+2(1600u2)x
(1300u2)x=u2
x=300 m
Hence total distance that the man can cover =300 m

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