Question

a particle moves on a vertical line so that its coordinate time t is s(t)=2t^3-18t+7, t>/=0
a. Find the velocity and acceleration functions
b. when the particle is moving upward and when it is moving downward?
c. find the instantaneous velocity at t=4

Answer 1

velocity is the derivative of s(t)
acceleration is the derivative of velocity... also second derivative of s(t)....

v(t) = s'(t)
a(t) = v'(t) = s''(t)

Answer 2

I have been doing ones that are like this but not set up like this.. so I'm lost

Answer 3

to get your velocity, just take the derivative of s(t).... what u got?

Answer 4

6t^2-18

Answer 5

so v(t) = 6t^2 - 18... now find a(t)...

Answer 6

12t

Answer 7

that was easy... now b...

Answer 8

your velocity tells you if you're moving up or down...

Answer 9

so we need to first find where v(t) = 0

Answer 10

Positive velocity means it's oriented in the direction of increasing \(s\).

Answer 11

so plug in 0 to 6t^2-18?

Answer 12

uh... solve \(\large 0=6t^2-18 \) for t...

Answer 13

oh!
square root of 6?

Answer 14

no.. try again...

Answer 15

3=t^2

Answer 16

good.. so what's t=?

Answer 17

1.73

Answer 18

is that sqrt3 ?

Answer 19

yeah

Answer 20

ok... what about -sqrt3 ? why did you omit that one?

Answer 21

i thought you couldnt have the square root of a negative number?

Answer 22

no i not talking about sqrt(-3)... i mean -sqrt(3)

Answer 23

oh you mean -1.73?

Answer 24

yeah... that...^^^

Answer 25

For your viewing pleasure, if this helps.

Answer 26

ok.. because your question said only for \(\large t\ge 0 \)

Answer 27

yeah that is what the question asked for..

Answer 28

just wanted to make sure...:)

Answer 29

so velocity is zero at t = sqrt 3...

Answer 30

what does that mean?