what is the derivative of (-4+5t)e^-.3t?

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lpaine1331

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I need to know the derivative of the following expression and can't quite get it right:


The question reads: The displacement of a particle which moves along the s-axis is given by s=(-4+5t)e^-.3t, where s is in meter and t is in seconds. Determine the time at which the acceleration is zero.

So, I need to get the derivative of the expression to get the velocity, then derivative again to get the acceleration equation, then plug in the numbers. I can't seem to get the first derivative......

Been along time

thanks
 

This is a SITE for engineers and students .But students should at least make the effort makeing their own homework .The world has become highly competitive, there are a lot of students that study hard .Those will be the ones that will get the jobs .Is of no use to get a simple problem solved by somebody else.
 

Hey EJ,
Go a bit easy on the guy... I think he is a newbie introduced to kinematics...

By the way lipane... EJ is correct you have to put in more effort into your homework... I will just give you the hint because you said you have spent enough time on it...

Use the product rule (UV)' = U'V + UV' where U = (-4+5t) and V = e^-.3t.

~Kalyan.
 

kalyanram said:
Hey EJ,


Use the product rule (UV)' = U'V + UV' where U = (-4+5t) and V = e^-.3t.

~Kalyan.
Click to expand...

dis will surely solve the problem.
if still wont able then simply multiply both quantities with e and seperately solve it
 

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