A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t) = (t)In(5t). Find the acceleration of the particle when the velocity is first zero.
Help. I think the answer is 5e^2

Question

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t) = (t)In(5t). Find the acceleration of the particle when the velocity is first zero.
Help. I think the answer is 5e^2

anonymous · Answer

Velocity = derivative of position function.

Acceleration = derivative of velocity function = second derivative of position function.

So the rephrase the question: When is $s'(t)=0$? What's the value of $s''$ at this point?

anonymous · Answer

You have
$$s(t)=t\ln5t\implies s'(t)=\ln5t+1\implies s''(t)=\frac{1}{t}$$
Find $t$ such that $\ln5t+1=0$ and plug it into the second derivative - pretty straightforward.

How are you getting $5e^2$? You're almost right.

kaleidoscopicsink · Answer

Kinda guessing. Because it looked right. SO would that be just 5e then?

anonymous · Answer

Yes.

kaleidoscopicsink · Answer

Can you explain that to me a little bit?

anonymous · Answer

Which part? The actual computation, or why this is the way to do it?

kaleidoscopicsink · Answer

The actual computation. If that is no trouble.

anonymous · Answer

With $s(t)=t\ln5t$, compute the first derivative via product rule:
$$\begin{align*}\frac{\mathrm{d}s}{\mathrm{d}t}&=\frac{\mathrm{d}}{\mathrm{d}t}[t\ln5t]\[1ex]&=\frac{\mathrm{d}}{\mathrm{d}t}[t]\ln5t+t\frac{\mathrm{d}}{\mathrm{d}t}[\ln5t]\[1ex]&=\ln5t+t\frac{5}{5t}\[1ex]
&=\ln5t+1\end{align*}$$
Set equal to $0$ and solve for $t$:
$$\ln5t+1=0\implies 5t=\frac{1}{e}\implies t=\frac{1}{5e}$$
Compute the second derivative:
$$\begin{align*}\frac{\mathrm{d}^2s}{\mathrm{d}t^2}&=\frac{\mathrm{d}}{\mathrm{d}t}[\ln5t+1]\[1ex]
&=\frac{5}{5t}\[1ex]
&=\frac{1}{t}\end{align*}$$
And finally, plug in $t=\dfrac{1}{5e}$, so the value of the second derivative is simply $5e$.

kaleidoscopicsink · Answer

Thank you so much. Can I have your assistance for one more question similar to this?

anonymous · Answer

You're welcome!

I'm actually gearing up for a 4 hour drive in the next few minutes, so I'll be indisposed. There are plenty of other users with a calc proficiency, and calc questions tend to have a pretty high rate of getting a timely response.

kaleidoscopicsink · Answer

Thank you for your help. Have fun with your drive. cx