# 2 particles moving along x axis.

1. Sep 27, 2007

### ~christina~

[SOLVED] 2 particles moving along x axis.

1. The problem statement, all variables and given/known data
Two particles move along the x axis. The position of particle 1 is given by

x= (6.00m/s^2)t^2 + (3.0m/s)t + 2.00m

the acceleration of particle 2 is given by:
a = - (8.00 m / s^3) t and, at t = 0, its velocity is 20 m / s.

(a) Plot the velocities of the two particles versus time on a single graph.
(b) Is either of the particles instantaneously at rest between t = 0 and t = 3s?
(b) is yes, then name the particle(s) and the time(s).
(d) Can the UAM equations be used to describe the motion of particle 2? Use the correct physics principles to support your response.
(e) Plot the accelerations of the two particles versus time on a single graph for the time interval given above.

2. Relevant equations
a = - (8.00 m / s3) t and, at t = 0, its velocity is 20 m / s.

3. The attempt at a solution
I want to check my thinking first of all since this a long problem.

Do I have to get the derivative of the equation of the equation given?

and what do I do with the values given for particle b?

How do I put them together to graph it??

for d.) what is a UAM equation???

~Thanks~

2. Sep 27, 2007

### MathematicalPhysicist

1. yes, for the velocities.
2. with the values given for the second particle you can find by integration the velocity of the body.
3.well the equations for velcoities give you the answer how.
4.im not sure about UAM, but perhaps it's uninform acceleration movement, although from the given you can see that the accleration isn't unifrom cause it changes with time.

3. Sep 27, 2007

### Avodyne

You are given x(t) for particle 1, and a(t) and v(0) for particle two. You need to figure out v(t) and a(t) for both. So, for particle 1 how do you get v(t) from x(t), and then a(t) from v(t)? For particle 2, how do you get v(t) from a(t) and v(0)?

4. Sep 28, 2007

### ~christina~

Is this alright?

for particle 1 :
v'= x(t)dx/dt = (6.00m/s^2)t^2 + (3.00m/s)t + 2.00m= (12m/s^2)*t + 3.0m/s

particle 2:
a= -(8.00m/s^3)t

v=$$\int a(t)$$= $$\int -(8.00m/s^3)t$$= -(4.00m/s)t^2 + t

~I don't get why they say that when t= 0 v= 20m/s for particle 2 since if I plug it into a equation it doesn't equal 20m/s

a.) well if I graph that direclty wouldn't it be wrong since they say the particle is moving at 20m/s at t= 0 ???

b.) For this part of the problem I don't know how to see whether the particles are instantaneously at rest since I put the equations into my calculator and saw that the first one is a parabola but the top flat part starts out at 0 and encompases quadrants 3 and 4. The other one is a straight line that starts in quadrant 3 and goes through quadrant 2.

~ I also don't know how to get the equation specifically for the velocity for both since they include units and I'm not sure how m/s => ? when derived or vice versa since I've always just worked with numbers and not actual units and the units were plugged in at the end.

Help..~Thanks~

5. Sep 28, 2007

### MathematicalPhysicist

first of all accleration is measured by meters per seconds^2.

now i see you don't know how to integrate, the indefinite integral of t wrt t is: t^2/2+c where c is a constant, form you are given you get v_2(t)=-4t^2+c
now if you plug in v_2(0)=-4*0^2+c you get that c=v_2(0) which is 20 and then you replug it to the equation.

6. Sep 28, 2007

### ~christina~

So I would get making the correction:

Particle 1: v(t) = x'(t)= x(t)dx/dt = (6.00m/s^2)t^2 + (3.00m/s)t + 2.00m= (12m/s)*t + 3.0m
(I corrected the units for particle 1 as well)

Particle 2: v(t)= $$\int ax(t) dt$$=$$\int - (8.00 m / s^3) t$$ = -(4.00m/s^2)t^2 + 20

The point you make about the acceleration being m/s^2 is something I already know but the question was given by my teacher as s^3 so I'm not sure if it's a typo or not...is velocity ALWAYS m/s^2? b/c that would mean that it's a typo and my v for the second particle would be v(t)= -(4.00m/s)t^2

(a) Plot the velocities of the two particles versus time on a single graph.
I did that and they intersect at x= 1.05, y= 15.59

(b) Is either of the particles instantaneously at rest between t = 0 and t = 3s?
I think that since the first particle is a straight line and keeps going up that it isn't instantaneously at rest at any point. Particle 2 however does level off at the top of the parabola so it would be yes

(c) If the answer to is yes, then name the particle(s) and the time(s).
I say it is particle 2 and the time is at t=20s

(d) Can the UAM equations be used to describe the motion of particle 2? Use the correct physics principles to support your response.
You are correct in saying it was the uniform accelerated motion equations but it DOES change with time but I guess but it doesn't count for the instantaneous part of the graph where the v= 0 right?

(e) Plot the accelerations of the two particles versus time on a single graph for the time interval given above.
Wouldn't a/t= J ?
Would I derive the equation of both equations given to get the acceleration for the time interval?

~THANKS~