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 -0.4t# Find the time at which the acceleration of the particle is greatest?

Question

2608 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-16T08:52:04

See below.

We have the velocity function:

#v=0.01t^3+0.22t-0.4t#

#"acceleration"=("change in velocity")/("time")#

Hence:

#a=(dv)/(dt)#

#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.

#(da)/(dt)(0.03t^2+.44t-0.4)=0.06t+0.44#

#0.06t+0.44=0=>t=-7.33s#

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