A particle moves along X axis. The velocity of the particle as function of x is v^2 = 12x - 3x^3 ( in SI ) . Find velocity of particle at the instant when acceleration is 1.5ms^-2
Answer (1)
22nd Feb, 2020
Hey aspirant,
according to the given data -
differentiate v^2 = 12x - 3x on both sides we get,
Therefore 2vdv=9dx
Therefore dv/dx=9/2v
Therefore a=9/2v (Since dv/dx=acceleration)
Therefore v=9/2a
Therefore v=3m/s (putting a=1.5 m/s^2)
Hope this helped,
Thank You.
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