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Question

# The potential energy of a particle of mass 5 kg moving in x-y plane is given as U=7x+24y joule, x and y being in metre. Initially at t=0, the particle is at the origin (0, 0) moving with a velocity of (8.6^i+23.2^j) ms−1. Then:

A
the velocity of the particle at t=4 s, is 5 ms1
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B
the acceleration of the particle is 5 ms2
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C
the direction of motion of the particle initially (at t=0) is at right angles to the distance of acceleration
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D
the path of the particle is circle
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Solution

## The correct options are A the velocity of the particle at t=4 s, is 5 ms−1 B the acceleration of the particle is 5 ms−2→F=−[δUδY^i+δUδy^j]=(−7^i−24^j)N→a=→Fm=(−75^i−245^j) m/s2∣^a∣=√(75)2+(245)2=5 m/s2Since, →a= constant, we can apply→v=→u+→at=(8.6^i+23.2^j)+(−75^i−245^j)(4) s=(3^i+4^j) m/s∣→v∣=√(3)3+(4)2=5 m/s

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