Question

A particle moves along the x axis. Its position is given by the equation
x = 1.9 + 2.5t − 3.7t2 with x in meters and t in seconds.

(a) Determine its position when it changes direction.
____m

(b) Determine its velocity when it returns to the position it had at t = 0? (Indicate the direction of the velocity with the sign of your answer.)
____m/s

Answer 1

i know the answer isnt 3.12 or 2.5

Answer 2

For part two of the question, when t = 0, x = +1.9

Answer 3

For part a)
find vertex
x = -b/2a

Answer 4

haha for this problem i should say --> t = -b/2a

Answer 5

ok

Answer 6

for a did u get 4.63?

Answer 7

so something like that?

Answer 8

no i get 2.32

--> -2.5/-7.4 = .3378
plug that in

Answer 9

its -2.5/2(-3.7) right?

Answer 10

yes

Answer 11

now i got what u got lol

Answer 12

are you given velocity function?
you can differentiate to get velocity but is this a calculus problem?

Answer 13

how did u get 2.32 then?

Answer 14

same way you did:)

no i am looking at part b

Answer 15

im confused lol what part were u doing lol

Answer 16

part a, position when particle changes direction

Answer 17

thats 2.32?

Answer 18

velocity is 2.5-7.2t. That is simply the first derivative of the position function, which is given to you. We know when it is changing direction its velocity is zero, so solve for t when velocity is 0.

Answer 19

then plug that t into the first equation to get the position.

Answer 20

yeah the first part should be....x = 2.32

velocity is derivative of x(t)
x'(t) = 2.5 - 7.4t
Because x(t) is parabola it is symmetric around t = .3378, therefore it will be in the same position .3378 sec after it changes direction.
t = 2*.3378 = .67567
plug that into x'(t) to find velocity

Answer 21

lets see if we get the same answer this time lol
i got for part b -2.5

Answer 22

yep we agree

Answer 23

yea its 7.4 not 7.2 mi bad
for the second part, when it starts the time is equal to 0, Plug t = 0 into the position function to get the position at t = 0.

Now that we know the position when it starts (and ends) is 1.9, plug that into the position function to find out the time when it ends.

Now that we know the time when it ends, plug that into the velocity function (which dumbcow posted) to find out its velocity when it ends.

Answer 24

yay!!!!! its right!!! thanks!!!!

Answer 25

stay tuned...possible new question will be posted =)

Answer 26

haha ok