Question

can you factor out 2t^2-9t-12 ?

Answer 1

no you cannot since there are no two numbers that multiply to 2(-12) = -24 AND that add to -9

Answer 2

use the quadratic formula.

Answer 3

okay i thought so...what math are you in ?

Answer 4

Calc 1

Answer 5

have you learned higher derivatives and the applications of the derivatives to velocity, acceleration, and rates of change ?

Answer 6

Yeah i have

Answer 7

Im not good at change of rate problems though :(

Answer 8

rate of change**

Answer 9

darn -.-

Answer 10

I hate those with a passion lol.. i know how to find velocity and acceleration though.

Answer 11

often, drawing a pic (a good one) really helps

Answer 12

would you like to help me ?..possibly ?

Answer 13

since you can lay out what the problem is giving you and what you want

Answer 14

sure

Answer 15

Velocity is the first derivative and acceleration is the second derivative

Answer 16

the position function of a particle is given by s=t^3-4.5t^2-7t where t is greater than or equal to 0
a. when does the particle reach a velocity of 5 m/s
b. when is the velocity 0
c. when is the acceleration 0

Answer 17

the velocity function is simply the derivative of the position function

Answer 18

so to answer parts a) and b), we need the velocity function

Answer 19

so what do you get when you derive the position function s(t)?

Answer 20

2t^2-9t-7

Answer 21

good

Answer 22

part a) when does the particle reach a velocity of 5 m/s


This is just asking for what values of t make the velocity function equal to 5


Mathematically, they want to know which t values satisfy the equation


v(t) = 5


But you found that v(t) = 2t^2-9t-7


So you're basically solving


2t^2-9t-7 = 5


for t

Answer 23

2t^2-9t-7 = 5


2t^2-9t-7 - 5 = 0


2t^2-9t-12=0


You can't factor this, but we can solve using the quadratic formula


t = (9+-sqrt(9^2-4(2)(-12)))/(2*2)


t = (9+-sqrt(177))/(4)


t = (9+sqrt(177))/(4) or t = (9-sqrt(177))/(4)


t = 5.576 or t = -1.076


Throw out the negative solution


So the only solution is t = 5.576 (this is a rough approximation of course)



So the velocity is 5 m/s at approximately 5.576 seconds

Answer 24

do the same for part b), but instead of using 5, use 0

Answer 25

thank youu (:

Answer 26

yw

Answer 27

And then for acceleration you find the derivative of the velocity equation and set it equal to 0 and solve.

Answer 28

okayy(: thank youu !