Question

calc. help

Answer 1

Yes???

Answer 2

@Mrs.ambrose614

Answer 3

anybody??

Answer 4

Yes you are

Answer 5

If:
\[s(t)=60t-1.5t^2\]
s(t)= position of car
t= seconds after driver hits breaks
\[s'(t)=60-2.25t\]
*Velocity --> when velocity = 0 car stopped
Find Critical Points:
\[2.25t=60\]
\[t=26.6\]
aka the car comes to a stop @ 26.6 seconds

Answer 6

(1.5t^2)'
=
3t

Answer 7

cool, thank you!!! WHAt about these 2?? Im really confused on both..

Answer 8

oh shoot lol @myininaya XD yeah computational error

Answer 9

@sunnnystrong @myininaya

Answer 10

it happens to us all :)

Answer 11

\[s'(t)=60-3t\]****
so/// Critical points @ t = 20sec

^^ same logic (:

Answer 12

@myininaya if you could please help me with the 2 i just posted, that would be fabulous!

Answer 13

@sunnnystrong you too :-) ^^

Answer 14

you mean 3?

Answer 15

question 3 and 4 are tagged in the same picture so, no just the 2^

Answer 16

do you need me to attach it again??

Answer 17

@iwanttogotostanford i will try lol

Answer 18

thank you so much!

Answer 19

are you asking help on finding the second derivative of t ln(2t) ?
I don't see question 2.

Answer 20

http://assets.openstudy.com/updates/attachments/584f1ec1e4b0d683d62e22ba-iwanttogotostanford-1481580801050-screenshot20161212at5.12.58pm.png

Answer 21

they are both right there in that link to the attachment i already attached above

Answer 22

You have two steps for number 3:
Solve s'(t)=0 for t
then plug that into s''(t)

Answer 23

do you know how to find s'(t)
if s(t)=t ln(2t)

you need product rule and chain rule

Answer 24

Okay so: for question 3
s(t)=position of particle
t=time (sec)
\[s(t)=t*\ln(2t)\]
recall that:
first derivative = velocity
second derivative = acceleration
\[s'(t)=(\frac{ 2 }{ 2t }*t)+(ln(2t))\]
\[s'(t)=ln(2t)+1\]

Answer 25

so... what @myininaya said. question 3 has 2 parts:
find when the velocity of the particle = 0 (aka critical point of first derivative)
& find the acceleration of the particle @ that time...

solving for critical point of first derivative:
*skipping some steps haha*
\[-1=\ln(2t)\]
\[e^{-1}=2t\]
\[t=\frac{ 1 }{ 2e }\]

Answer 26

\[s''(t)=\frac{ 1 }{ t }\]

Compute s''(t) @ t=2e^(-1)
\[s''(2e ^{-1})=\frac{ 1 }{ 2e ^{-1} }\]
=2e