Physics particle velocity kinematics question

[b-pipe]

Homework Statement

At t=0, a particle is located at x=33m and has a velocity of 19 m/s in the positive X direction. The acceleration of the particle varies with time as shown in the diagram. What is the velocity of the particle at t=3.6s

Diagram:

Homework Equations

I'm not sure. But I assume its a variation on the equations of motion.
http://en.wikipedia.org/wiki/Equations_of_motion#Equations_of_uniformly_accelerated_linear_motion

The Attempt at a Solution

Ive gotten the variables:

Initial Position: 33m
Initial Velocity: 19m/s
Initial Time = 0s
Final Time = 3.6 m/s
Acceleration = Unknown

I initially tried to plug the numbers into one of the equations of motion but couldn't come up with an answer. I realized that the equations of motion are only applicable if the acceleration is constant, which the question states it is not. This is where I'm at a loss. I also think that the diagram provides some information on the acceleration but I'm not sure how I'm supposed to get the info from the graph depicted.

Any help would be much appreciated.

 

[PhanthomJay]

Are you familiar with the calculus, A=dv/dt? First find the equation of the graph ("A" as a function of "t").

 

[b-pipe]

Are you familiar with the calculus, A=dv/dt? First find the equation of the graph ("A" as a function of "t").

Yes, I am. In my calculus class we were taught that. So I should find the equation of the graph? I'm a bit rusty so please give me a minute.

Edit: The equation of the line is y= - 6/5x + 6

 

[gneill]

Acceleration is the derivative of velocity, and so naturally, velocity is the integral of acceleration. You've a nice graph there where area under the curve is pretty easy to find...

 

[b-pipe]

Acceleration is the derivative of velocity, and so naturally, velocity is the integral of acceleration. You've a nice graph there where area under the curve is pretty easy to find...

I'm confused at what equation of velocity I should be integrating. Should it be the equation for the line of the graph I posted? Did I give the correct equation in my edit?

 

[gneill]

I'm confused at what equation of velocity I should be integrating. Should it be the equation for the line of the graph I posted? Did I give the correct equation in my edit?

If you integrate acceleration you get velocity. You have a graph of the acceleration. How do find the area under a curve?

Yes, your equation of the line is fine. You might want to use appropriate variable names and units to make things pretty. a(t) = 6m/s² - (6/5)(m/s³)*t .

You can integrate a(t) to find the velocity (or calculate it directly from the graph with a bit of geometry). Don't forget that the particle already has an initial velocity!

 

[b-pipe]

If you integrate acceleration you get velocity. You have a graph of the acceleration. How do find the area under a curve?

Yes, your equation of the line is fine. You might want to use appropriate variable names and units to make things pretty. a(t) = 6m/s² - (6/5)(m/s³)*t .

You can integrate a(t) to find the velocity (or calculate it directly from the graph with a bit of geometry). Don't forget that the particle already has an initial velocity!

Thank you very much.

 

[b-pipe]

Hello again,

I'm at a bit of a problem. When I intergrate the equation y= - 6/5x + 6, I get y = -3/5*x^2 + 6x. If I evaluate the resulting equation from 0 to 5 will the answer be the velocity?

 

[gneill]

Hello again,

I'm at a bit of a problem. When I intergrate the equation y= - 6/5x + 6, I get y = -3/5*x^2 + 6x. If I evaluate the resulting equation from 0 to 5 will the answer be the velocity?

It would be the velocity (added to the initial velocity at t = 0) of the particle at t = 5s.

You want the the velocity at t = 3.6s.