Question

# Instantaneous velocity of a particle moving in a straight line is given as v=(4+4√t) m/s, for the first 5 sec of its motion, there after velocity becomes constant. Find the acceleration and displacement of the particle at time t=3.0 sec.

A
23 m/s2, 35.50m
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B
23 m/s2, 25.85m
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C
5 m/s2, 25.85m
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D
10 m/s2, 15.85m
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Solution

## The correct option is B 2√3 m/s2, 25.85mGiven, velocity of particle for the first 5 sec, v=(4+4√t) m/s As we know that acceleration of a particle is time derivative of its instantaneous velocity.a=dvdt ⇒a=d(4+4√t)dt ⇒a=2√t And at time t=3.0 sec, acceleration is a=2√3 m/s2 To find displacement: We know that velocity is the time derivative of displacement, we have to integrate as follows v=dxdt=4+4√t ⇒dx=(4+4√t)dt Integrating the above expression from time t=0 to t=3.0 sec, we get the displacement, which varies from xi=0 to xf=x, for these time instants. ∫xf=xxi=0dx=∫t=3t=0(4+4√t)dt ⇒[x−0]=[4t+83t3/2]30 ⇒x=[12+13.85] ∴x≈25.85 m Hence, option (b) is the correct answer.

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