Question

a couple of calculus questions regarding velocity, and position, acceleration type of stuff

Answer 1

1. A particle is moving along the x-axis so that at time t its acceleration is . At time t = ½, the velocity, v of the particle is ½.

a) Find the velocity of the particle at any time t.
b) Find the minimum velocity of the particle.
c) Find the equation for the position x(t) if x(0) = 0.
d) What is the first time t > 0 that the particle returns to the origin?




2. A particle is moving with velocity:



with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t = 0 sec is 1 meter right of zero.

Find each of the following. Give units in the answer.

a) The initial acceleration
b) The average velocity over the interval 0 to 8 seconds
c) The instantaneous velocity at time 5 secs
d) The time(s) when the particle is at rest
e) The time interval(s) when the particle is moving left
f) The time interval(s) when the particle is moving right
g) The speed of the particle at time 4 secs
h) The time interval(s) when the particle is
(i) going faster
(ii) slowing down
i) Find the total distance the particle has traveled between 0 and 8 seconds

Answer 2

what is the rate of acceleration?

Answer 3

i dont know, this is all the info i have.

Answer 4

1. A particle is moving along the x-axis so that `at time t its acceleration is ` . At time t = ½, the velocity, v of the particle is ½.

Are you sure? it looks like something didn't paste correctly.

Answer 5

ah
your right.

a(t) = picospi

Answer 6

1/2 velocity is incorrect description
velocity is a vector so it needs to have both magnitude and direction

Answer 7

i dunno why it didnt copy tho.

Answer 8

you're not allowed to copy and paste your homework here btw

Answer 9

its my review questions for my test.

Answer 10

and im confused.

Answer 11

alright, i dont mind that

Answer 12

but if you want the file, i can attach it. for proof

Answer 13

So the acceleration is given to be constant.
Have you been introduced to integration yet?
Or just the idea of anti-differentiation?

Answer 14

yes, i have

Answer 15

\[\Large a(t)\quad=\quad \pi \cos \pi\]Taking the anti-derivative of acceleration should give us the velocity function.
Do you understand why that is?
Do you know how to find the anti-derivative of this function? :)

*cough* anti-derivative of a constant? *cough*

Answer 16

because the derivative of velocity would equal the acceleration?

and the anti-derivative would be -pix

Answer 17

Well we wouldn't be getting an x, we'd get a t in this case.
But yes! You have the right idea.
We'd also want a constant of integration.\[\Large\bf\sf v(t)\quad=\quad -\pi t+c\]

Answer 18

ah right, forgot about that:P

Answer 19

We can use our initial data to solve for the unknown constant.\[\Large\bf\sf v\left(\frac{1}{2}\right)\quad=\quad \frac{1}{2}\]We'll plug this into our velocity function to solve for c.

Answer 20

so 1/2 = pi(1/2) + c
so c= (pi)(1/2) - 1/2)

Answer 21

which is equal to 1.0708

Answer 22

we had a negative in front of our pi didn't we?

1/2 = -pi(1/2) + c
so c= pi(1/2) + 1/2

yes? :o

Answer 23

ah, so it would be 2.0708 instead

Answer 24

So for part a, we want to velocity function with our constant plugged in.\[\Large\bf\sf v(t)\quad=\quad -\pi t+\color{orangered}{c}\]

Answer 25

want the*

Answer 26

It probably looks nicer if we don't approximate.\[\Large\bf\sf v(t)\quad=\quad -\pi t+\color{orangered}{\frac{1}{2}\pi+\frac{1}{2}}\]But do whatever makes more sense to you.

Answer 27

so the answer to a is) v(t) = -pi(t) + 2.0708

Answer 28

ya looks good c:

Answer 29

how would i fine the minimum velocity, by plugging in 0?

Answer 30

If the acceleration is `constant` then the velocity is `linear`.

Since the velocity function has a negative coefficient in front, it will decrease at a constant rate.

The velocity function won't have any critical point which we could call a minimum.

If you wanted to be a little more rigorous you could take the derivative of v(t) and then set it equal to zero and solve for t to find critical points.

Answer 31

\[\Large 0\quad=\quad \pi \cos \pi\]No t to solve for :c
No critical points.
No minimum.

Answer 32

could the minimum velocity be said to be -infinity?

Answer 33

Ummmm no I guess we wouldn't want to do that.
A minimum should be an actual value.

I think... -_-

Answer 34

so how would i find the position formula

Answer 35

Do you remember how position relates to velocity? :)
*hint hint* similarly to the way velocity relates to acceleration!

Answer 36

i should take the anti-derivative of v(t)?

so that would give me (-pix^2)/2 + c

Answer 37

since s(0) is 0, c is 0, right?

Answer 38

Hmm what happened to your previous c in v(t)?

Answer 39

Let's call our new constant d or something, to avoid confusion.

Answer 40

er wait, i just took the antiderivative of the original >.>

Answer 41

\[\Large\bf\sf v(t)\quad=\quad -\pi t+\left(\frac{1}{2}\pi+\frac{1}{2}\right)\]

Answer 42

So we want to anti-differentiate that.

Answer 43

so it would be x(1/2pi + 1/2) + x(1/2pix)

Answer 44

+ c. but c is 0 because of s(0) is equal to 0

Answer 45

Ok looks like you're on the right track.
Where did our negative again though? :(
\[\Large\bf\sf s(t)\quad=\quad -\frac{1}{2}\pi t^2+\left(\frac{1}{2}\pi+\frac{1}{2}\right)t+d\]

Answer 46

s(0)=0
Oh ok good good c=0

Answer 47

er, its (.5pi + .5) - x(.5pix)

Answer 48

ok cool so that takes care of part c! :)

Answer 49

i dont understand part d at all

Answer 50

So if you look at your position function a sec.
What shape does it take?

The largest power on x is a square, and it has a negative coefficient.
Can you recall what shape it will make? :o

Answer 51

a bunch of u's.

when i graphed the position, it was a linear line, almost like y=x

Answer 52

Hmm it shouldn't give you that..

It should give you one u.
It's a parabola opening downward.

Answer 53

o, i graphed it wrong:P

Answer 54

part d is simply asking you to `find the x-intercepts`.

They're calling the x-axis the "origin" in this case, which is a little strange.
But I guess it makes sense.

They want to know when the particle touches back down to the ground or whatever.

Answer 55

Remember how to find `x-intercepts`? Also called `zeros` or `roots` of a polynomial.

Answer 56

zeros are 0 and 1.318

Answer 57

Ok good. and it looks like part d is telling us to ignore one of those points, yes?

Answer 58

Since one of those points is the starting location.

Answer 59

so 1.318 is our answer

Answer 60

At time t = 1.318 seconds, the particle will return to the origin.

Ya I think that sounds right! :)

Answer 61

what about number 2, whats initial acceleration?

Answer 62

Ahh I dunno. I'm too tired. I need to get some sleep :c
Maybe mr poo can help you finish this one up.

Answer 63

aite, tyvm.

Answer 64

People hate to read really long posted questions.

It might be worth your time to close this thread and open a new question with just #2.
You might be able to find some help easier that way c: