A particle of mass 2/3 kg with velocity v = –15 m/s at t = –2 s is acted upon by a force \f = k -\betat^2 . Here k = 8N and β = 2N/s^{2} . The motion is one dimensional. Then the speed at which the particle acceleration is zero again, is

KVPY 2019 SB SX Exam Previous Papers

Question : 27 of 120

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A particle of mass 2/3 kg with velocity v = –15 m/s at t = –2 s is acted upon by a force

\f = k -βt^2

. Here k = 8N and β = 2N/s

^{2}

. The motion is one dimensional. Then the speed at which the particle acceleration is zero again, is

Solution:

\f = 8 -2t^2 =m{dv}/{dt}

2/3{dv}/{dt}=8-2t^2

∫dv=3/2∫(8-2t^2)dt

⇒

v=3/2[8t-{2t^2}/3+C]

......(1)
On putting at t = –2 sec ,v = –15 sec
⇒

C = 2/3

F is zero again at t = 2 sec putting t in (A)
So v' = 17 m/s

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