a velocity of particle moving along x axis it's given as v=x^2-5x+4 where x denotes the x co orddinate of particle in meters . find magnitiude of acceleretion pf the particle when velocity is zero?
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When v =0
v = x²-5x+4 = (x-1)(x-4)
Therefore value of v is zero at x = 1 and x= 4.
acceleration = dv/dt = d(x²-5x+4)/dt
= d(x²-5x+4)/dx × dx/dt
= (2x-5)×v
= (2x-5)×(x²-5x+4)
Therefore, when v= 0, a = 0.
I hope you understand the approach.
All the best!
v = x²-5x+4 = (x-1)(x-4)
Therefore value of v is zero at x = 1 and x= 4.
acceleration = dv/dt = d(x²-5x+4)/dt
= d(x²-5x+4)/dx × dx/dt
= (2x-5)×v
= (2x-5)×(x²-5x+4)
Therefore, when v= 0, a = 0.
I hope you understand the approach.
All the best!
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