A particle starts from rest and moved in a straight line the velocity of the particle at time t after start is v m/s where $v = 0.01 t^{3} + 0.22 t^{2} - 0.4 t$ $v= 0.01t^3 +0.22t^2 -0.4t$ Find the time at which the acceleration of the particle is greatest?

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Answer 1 · 2018-05-16T08:52:04

See below.

We have the velocity function:

$v = 0.01 t^{3} + 0.22 t - 0.4 t$ $v=0.01t^3+0.22t-0.4t$

$acceleration = \frac{change in velocity}{time}$ $"acceleration"=("change in velocity")/("time")$

Hence:

$a = \frac{d v}{d t}$ $a=(dv)/(dt)$

$a = \frac{d v}{d t} (0.01 t^{3} + 0.22 t - 0.4 t) = 0.03 t^{2} + 0.44 t - 0.4$ $a=(dv)/(dt)(0.01t^3+0.22t-0.4t)=0.03t^2+0.44t-0.4$

This is the function of acceleration.

We need to find maximum value of this. Using the derivative of this and equating to zero will identify local maxima and minima.

$\frac{d a}{d t} (0.03 t^{2} + .44 t - 0.4) = 0.06 t + 0.44$ $(da)/(dt)(0.03t^2+.44t-0.4)=0.06t+0.44$

$0.06 t + 0.44 = 0 \Rightarrow t = - 7.33 s$ $0.06t+0.44=0=>t=-7.33s$

As pointed out in the comments, there must be an error in the function.

A particle starts from rest and moved in a straight line the velocity of the particle at time t after start is v m/s where v= 0.01t^3 +0.22t^2 -0.4tv=0.01t3+0.22t2−0.4tv= 0.01t^3 +0.22t^2 -0.4t Find the time at which the acceleration of the particle is greatest?