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
$a=0$