Question

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t) = (t)In(3t). Find the acceleration of the particle when the velocity is first zero.

Answer 1

You first need to find the point where the velocity of the particle is zero, and then to find the acceleration at this point (when the velocity is 0).

Answer 2

The position is:
\(\color{#000000 }{ \displaystyle s(t)=t\ln (3t) }\)

Answer 3

The slope of the position graph tells you how fast
the function is going. Or, s'(t) would tell you how
fast the particle is moving at each instant "t".

Answer 4

So you need to find the derivative:
\(\color{#000000 }{ \displaystyle s'(t)=}\)

(Use the product rule)

Answer 5

Ummm I don't know @solomonzelman

Answer 6

Be specific please. What don't you know?

Answer 7

Would the derivative be ln(3t)+1?

Answer 8

@solomonzelman

Answer 9

\(\color{#000000 }{ \displaystyle s'(t)=t\times \left(\ln(3t)\right)'+\left(t\right)'\times \ln(3t)}\)

\(\color{#000000 }{ \displaystyle s'(t)=t\times \left(\frac{1}{3t}\times 3\right)+1\times \ln(3t)}\)

\(\color{#000000 }{ \displaystyle s'(t)=1+ \ln(3t)}\)

Answer 10

Very good!

Answer 11

s'(t) is the velocity, so you can say,

v(t)=1+ln(3t)

Answer 12

Now, when is the velocity of the particle 0?
(at what values of t?)

Answer 13

to answer this question set v(t)=0

Answer 14

Ok thanks

Answer 15

And that is not going to be the end.

Answer 16

Can you solve

0=ln(3t)+1 ?

Answer 17

I know you solve it thanks that's all the help I needed

Answer 18

Just to make sure \( ...\)

After you solve the above equation for t,
you will need to find the acceleration at
this point.

The derivative of velocity is acceleration,
so v'(t) , (or s''(t)).

Find s''(t), and then plug in that value of t,
at which s'(t) (or v(t)) =0.