Particle Problem Inst. Velocity=Avg. Velocity?

[Loppyfoot]

Homework Statement

A particle moves along the x-axis so that at time t≥0 its position is given by:

x(t)= t^3-2t^2-4t+6.

For what value(s) of t where 0≤t≤4 is the particle's instantaneous velocity the same as its average velocity on [0,4]?

The Attempt at a Solution

So I got the average velocity:

v(4)-v(0)/ (4-0) = 22-6/(4-0) = 4 = avg velocity for v(t)

So then would I solve the derivative of the position function to find what value of t in the derivative would make x'(t)= 4?

Thanks in advance.

 

[Dick]

That's exactly what you should do.

 

[Loppyfoot]

so x'(t)= 3t^2-4t-4

so 3t^2-4t-4 = 4

= 3t^2-4t-8=0

Solve for t using the quadratic formula, where I get a positive and negative number. the positive number would be the only number that could fit on this interval. Would that be all?

Is there anything else I have to do with this problem?

Thanks

 

[HallsofIvy]

Homework Statement

A particle moves along the x-axis so that at time t≥0 its position is given by:

x(t)= t^3-2t^2-4t+6.

For what value(s) of t where 0≤t≤4 is the particle's instantaneous velocity the same as its average velocity on [0,4]?

The Attempt at a Solution

So I got the average velocity:

v(4)-v(0)/ (4-0) = 22-6/(4-0) = 4 = avg velocity for v(t)

So then would I solve the derivative of the position function to find what value of t in the derivative would make x'(t)= 4?

Thanks in advance.

But why did you change from "x(t)" to "v(t)" for the postion function?

 

[Loppyfoot]

Oops. the average velocity should be x(t) not v(t). Just got caught up with the "v"'s.

 

[Dick]

so x'(t)= 3t^2-4t-4

so 3t^2-4t-4 = 4

= 3t^2-4t-8=0

Solve for t using the quadratic formula, where I get a positive and negative number. the positive number would be the only number that could fit on this interval. Would that be all?

Is there anything else I have to do with this problem?

Thanks

That's all.