Question

A particle moves alone the x-axis so that its acceleration at any time t≥0 is given by a(t)=2t-6. The initial velocity of the particle is -27 and at time t=1, its position is 20.
For what values of t, 0≤t≤3, is the particle's instantaneous velocity the same as its average velocity on the closed interval [0,3]?

Answer 1

I has got
s(t) = (1/3)t^3 - 3t^2 - 27t + 149/3
and
v(t) = t^2 - 6t - 27

which meh teacher said was right, but idk what to do with themz now XD

Answer 2

oh and no calculator question... so gimme a while XD

Answer 3

Let's see...
\(a(t) = 2t - 6\)

\(v(t) = t^{2} - 6t + v_{0}\)

We are given \(v_{0} = -27\)

\(v(t) = t^{2} - 6t - 27\)

\(s(t) = (1/3)t^{3} - 3t^{2} - 27t + s_{0}\)

We are given: \(s(1) = (1/3) - 3 - 27 + s_{0} = 20 \implies s_{0} = (149/3)\)

\(s(t) = (1/3)t^{3} - 3t^{2} - 27t + 149/3\)

Very good. Now the question...

For what values of t, 0≤t≤3, is the particle's instantaneous velocity the same as its average velocity on the closed interval [0,3]?

How do we find the Average Velocity on [0,3]? \(\dfrac{v(3) - v(0)}{3-0}\)? What say you? Is that the average velocity or is it something else?

Answer 4

I did v(0) and v(3) tkhunny
but got it wrong. I'm guessing that was avg accel

Answer 5

so I has to do s(0) and s(3)
... s(3)...
0.o
no calculator...

Answer 6

@nincompoop do I get that avg velocity = 1/3 ? :3

Answer 7

Why did you do that? You didn't answer the question. Don't do it unless you KNOW what it is.

Average velocity is \(\dfrac{How\;far\;did\;we\;get?}{How\;long\;did\;it\;take?} = \dfrac{s(3)-s(0)}{3-0} = \dfrac{}{}\)

Okay, what do you get for Average Velocity?

Answer 8

1/3 ?

Answer 9

One piece at a time.

s(3) = ??
s(0) = ??

Please provide these values. No guessing.

Answer 10

152/3
and 149/3

so avg velocity = 1/3

Answer 11

Okay, s(3) isn't working. Please give that another go. Be more careful, this time.

Answer 12

(1/3)(3^3) - 3(3^2) - 27(3) + 149/3
(27/3) - 27 - 81 + 149/3
9 - 108 + 149/3
- 99 + 149/3
-297/3 + 149/3
-148/3 ?
so I did 99 times 3 = 307 wrong? XD no calculator sucks, I can't math >,<

Answer 13

That's better. So the average velocity is? (Don't forget to divide by (3-0)).

Answer 14

148+149 = 297
297/3 = 99
so -99?

Answer 15

You didn't read inside those last parentheses. \(\dfrac{s(3) - s(0)}{3 - 0}\). There is a Denominator, there.

Answer 16

oh, so 99/3 = 33
so -33?

Answer 17

Ding!!! Okay, off to part II. Find every time t in [0,3] where the Velocity actually takes on this value.

Answer 18

so then I look for -33 = (t-9)(t+3) ?

Answer 19

I suppose, but I have no idea why you bothered to factor it. Factoring does little good unless the other side of the equation is ZERO!

Answer 20

oh yeah...
so -33 = t^2 - 6t - 27
then t^2 - 6t + 6 = 0 ?

Answer 21

Okay, two more things to do.

Answer 22

no no no not quadratic equation, please not quadratic equation
i don't wanna D':
I had it programed into meh calculator and everything
so I wouldn't have to memorize it >,<

Answer 23

I'm hopeless aren't I >,<

Answer 24

Well, Complete the Square, then.

Answer 25

how would completeing the square help me?
(t-3)^2 -3 = 0

Answer 26

Where do you think the Quadratic Formula came from?

Solve for t.

Answer 27

oh...
(t-3)^2 = 3
then t-3 = √3 & -√3
t = 3 + √3 and 3 - √3 ?

Answer 28

Okay, now answer the question. What t on [0,3] have we just found?

Answer 29

oh 3 - √3

Answer 30

Done!

Answer 31

OMG THANK YOU!!! :D

Answer 32

Free Advice: Get that quadratic formula stuck in your head. Completing the square is fine, and that IS where the formula came from, but a little better speed and accuracy can be achieved with the formula well-memorized.

Answer 33

BTW Psyduck rules!!!

Answer 34

[-B +/- SQRT(B^2 -4AC)]2A

Answer 35

ahahahaha XD thanks tkhunny!

Answer 36

if you can remember discriminant you're halfway done