The velocity of a particle moving along x - axis is given as #v = x^2 - 5x + 4#(in m/s), where x denotes the x-coordinate of the particle in meters. Find the magnitude of acceleration of the particle when the velocity of particle is zero?

Question

A: #0m/s^2#
B: #2m/s^2#
C: #3m/s^2#
D: None of the above

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Answer 1 · 2017-08-08T14:49:05

A

Given velocity
#v=x^2−5x+4#
Acceleration #a-=(dv)/dt#

#:.a=d/dt(x^2−5x+4)#
#=>a=(2x(dx)/dt−5(dx)/dt)#

We also know that #(dx)/dt-=v#

#=>a=(2x −5)v#

at #v=0# above equation becomes
#a=0#