Question

Particles moving on vertical lines! Please check my work and help me with what to do next :)

Answer 1

Picture 1: original question
picture 2: my work work so far o.o

Answer 2

@ganeshie8 :)

Answer 3

@zepdrix ^-^

Answer 4

a is correct. Well done! (still looking through the others)

Answer 5

Thank you! I'll brb.

Answer 6

In part b, when you factor the polynomial, be careful of the signs in the terms. Double check that part.

Answer 7

hmm

Answer 8

wait how can I factor that o.o

Answer 9

without making the 6 positive

Answer 10

You'll have to use the quadratic formula. It doesn't factor.

Answer 11

For part (b), you will follow these steps


step 1) solve v(t) = 0 by using the quadratic formula


step 2) make a sign chart to figure out the intervals where v(t) > 0 (when the particle is moving upward) and where v(t) < 0 (when the particle is moving downward).


Keep in mind that t >= 0

Answer 12

Sorry, you guys! I had a phone call o.o but I am back now :)

Answer 13

Gah quadratic formula is too long ago -.- how do I do that?

Answer 14

Quadratic Formula
\[\Large x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Answer 15

In the case of v(t) = 2t^2 - 14t - 12, we have

a = 2
b = -14
c = -12

Answer 16

ok, give me a moment :)

Answer 17

(7- sqroot73)/2
(7+sqroot73)/2

Answer 18

correct

Answer 19

what are those values approximately equal to? in decimal form

Answer 20

- one = -0.77200
+ one = 7.77200

Answer 21

so we can ignore the root `t = -0.77200` since `t >= 0`

Answer 22

we only have to deal with `t = 7.77200`

now make a sign chart with 7.77200 on the number line and determine where v(t) is positive or negative

Answer 23

by choosing numbers on either side of it, yes?

Answer 24

yes

Answer 25

It's positive when it's larger than 77.772

Answer 26

I think you mean 7.772, but yes, correct

Answer 27

the object is moving upward when t > 7.772

Answer 28

and downward when t<7.772

Answer 29

when 0 <= t < 7.772

Answer 30

is that equals meant to be in there?

Answer 31

yes because \(\Large t \ge 0\)

Answer 32

so \(\Large 0 \le t < 7.772\)

Answer 33

Oh I see :) thanks.
So I guess that answers of all b!

Answer 34

To answer c?

Answer 35

I plug 0 into the original function y(x) to find where the particle begins?

Answer 36

let f(x) be the position function on a vertical number line

f(0) is the object's position at time t = 0 seconds
f(5) is the object's position at time t = 5 seconds

|f(5) - f(0)| is the total distance the object travels from t = 0 to t = 5 seconds

keep in mind that if the object turned around anywhere in the interval [0,5], then we'd have to break up the interval into smaller pieces. But the object doesn't turn until 7.772 seconds, which is outside the interval [0,5]

Answer 37

So in this case I can just plug 0 and 5 into the original function.

Answer 38

btw that's equivalent to saying \[\Large \int_0^5 v(t)dt\]

Answer 39

yes and subtract. Make the final result positive

Answer 40

Emm..
y(0)=12
y(5)=-139.667

Answer 41

both incorrect

Answer 42

Ai. Do I plug it into v(t)??

Answer 43

into the position function, not the velocity function

Answer 44

Yeah that's what I did and got the first values?
(2/3)(5)^3-7(5)^2-12(5)+12

Answer 45

I see `+2` at the end, not `+12`

Answer 46

oh gosh. I copied it down wrong on my paper.

Answer 47

after subtracting we get
151.6667

Answer 48

me too

Answer 49

Excellent. Sorry about the confusion. Part d!

Answer 50

So it speeds up when acceleration and velocity match signs (both - or both +)
and slows down when they are not matching. Yes?

Answer 51

Q: When is the particle speeding up? When is it slowing down?


A: If the velocity and acceleration are the same sign (both positive together or both negative together), then the object is speeding up. Otherwise, if the velocity and acceleration differ in sign, then the object is slowing down.


See the attached image chart

Answer 52

Yes you beat me to it

Answer 53

haha. So where do I test if acceleration is equal (Signs) to velocity?

Answer 54

the same as I did for finding when the particle was moving upward or downward?

Answer 55

I would make a sign chart for both v(t) and a(t). Line up the number lines (one over the other)

Answer 56

This is pretty much what i did for velocity. I can plug those random values into a(t) as well and see where the signs match?

Answer 57

this is how I did it (using geogebra and MS paint)

Answer 58

I found that a(t) = 0 leads to 4t - 14 = 0 which means t = 14/4 = 7/2 = 3.5

that's when a(t) flips from negative to positive

Answer 59

So it's speeding up at: (0,3.5)u(7.5,10)
and slowing down at (3.5,7.5)

Answer 60

where are you getting 7.5 ?

Answer 61

*7.772 I was just rounding. sorry.

Answer 62

since 10 is not a set number, should I put to infinity? or is that not safe to say?

Answer 63

oh gotcha, yes the object speeds up on the interval (0,3.5) U (7.772, infinity). The object slows down on the interval (3.5, 7.772)

Answer 64

It's so gorgeous *-*

Answer 65

why stop at 10? why not go off to infinity?

Answer 66

Right right.

Answer 67

Thank you so much

Answer 68

you're welcome