Finding maximum speed of v(t) = ate^(-3t)

[Minestra]

Homework Statement

The instantaneous speed of a particle moving along one straight line is v(t) = ate^-3t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)

2. The attempt at a solution
As far as I know this is a max min type problem you'd see in calculus. So I take the derivative and set it to zero,
v'(t)= -3ae^-3t
when I get to this point I kinda hit a wall because setting the problem to zero ultimately just gets me zero. I know the answer isn't zero. Any help with what I'm doing wrong will be much appreciated.

 

[haruspex]

Homework Statement

The instantaneous speed of a particle moving along one straight line is v(t) = ate^-3t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)

2. The attempt at a solution
As far as I know this is a max min type problem you'd see in calculus. So I take the derivative and set it to zero,
v'(t)= -3ae^-3t
when I get to this point I kinda hit a wall because setting the problem to zero ultimately just gets me zero. I know the answer isn't zero. Any help with what I'm doing wrong will be much appreciated.

You seem to have differentiated ae^-3t, not ate^-3t. Use the product rule.

 

[TomHart]

You may want to check your derivative.
Sorry I'm late. See post #2.

 

[Minestra]

You seem to have differentiated ae^-3t, not ate^-3t. Use the product rule.

Hazah, thank you. Its a rather embarrassing mistake when I see it now.