Question

"Given the position function of s of a point P moving on a coordinate line l, find the times at which the velocity is the given value k."

s(t)=3t^(2/3) ; k=4

The answer in the book says it is 1/8.

I am having trouble approaching this problem to get to the answer.

Answer 1

Basically the velocity is the derivative: \[
v(t) = s'(t)
\]And they want you to find the times (\(t\) values), where\[
v(t)=k \implies v(t)=4
\]

Answer 2

So would \[s'(t)=2t ^{-2/3}\]

Answer 3

First do \[
\frac 23-1 = \frac 23 - \frac 33=-\frac 13
\]

Answer 4

Oh I see so it's \[2t ^{-1/3}\]

Answer 5

You have to solve for \(t\):\[
2t^{-1/3}=4
\]

Answer 6

Ok so I see the derivative equal to 4 then solve for t. Now i stuck at
\[t ^{-1/3}=2\]
Hopefully I am solving this correctly.

Answer 7

set*

Answer 8

You aren't sure how to cancel out that exponent?

Answer 9

I kinda forgot

Answer 10

Need a little refresher

Answer 11

Try taking both sides by the natural log of 2. \[
\log_2(t^{-1/3})=\log_2(2)
\]

Answer 12

wouldn't that just be end up as \[t ^{-1/3}=2\]

Answer 13

It would give you \[
-\frac 13 \log_2(t)=1
\]

Answer 14

ooh so you pull out the exponent to the front?

Answer 15

Okay so when you simplify you get: \[
\log_2(t)=-3
\]

Answer 16

The anti log of both sides: \[
t=2^{-3}
\]

Answer 17

So, as you can see, from the beginning all you really had to do with take both sides to the power of \(-3\)

Answer 18

\[
(t^{-1/3})^{-3}=2^{-3}
\]

Answer 19

The reason I showed you the logarithm part is because it seems like your algebra is rusty and I wanted to teach logarithms.

Answer 20

oh! omg i remember that now, that was so simple.can't believe i forgot. thanks a lot! :D

Answer 21

yeah my algebra is verrrry rusty. i have to go back and review.