If velocity of the particle is given by v=√x, where x denotes the position of the particle and initially particle was at x=4m, then which of the following are correct? (Given velocity is in m/s and position is in m)
A
At t=2 sec, the position of the particle is at x=9m
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B
Particle's acceleration at t=2 sec. is 1ms2.
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C
Particle's acceleration is 12ms2 throughout the motion.
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D
Particle will never go in negative direction from it's starting position.
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Solution
The correct options are A At t=2 sec, the position of the particle is at x=9m C Particle's acceleration is 12ms2 throughout the motion. D Particle will never go in negative direction from it's starting position. Given, v=√x dxdt=√x ∫dx√x=∫dt 2√x=t+c at t=0,x=4 ⇒c=4 ⇒√x=t2+2
Position of the particle x=(t2+2)2 x=t24+2t+4............(1) x|t=2=1+4+4=9
Differentiating (1) v=t2+2................(2)
Differentiating (2) a=dvdt a=12ms−2 ( a is constant throughout the motion) Since t≥0⇒x≥0----from(1)