# Physics particle velocity kinematics question

• b-pipe
In summary, 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. At t=3.6s the particle has a velocity of unknown value.
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.

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

PhanthomJay said:
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

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...

gneill said:
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?

b-pipe said:
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!

gneill said:
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.

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?

b-pipe said:
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.

## What is the definition of particle velocity in physics?

Particle velocity is a measure of the rate at which a particle moves in a specific direction. It is a vector quantity that includes both the magnitude and direction of the particle's motion.

## How is particle velocity related to displacement?

Particle velocity is the rate of change of displacement. In other words, it is the derivative of displacement with respect to time. This means that the particle's velocity at any given moment is equal to the slope of its position-time graph at that moment.

## What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement of a particle divided by the total time taken, while instantaneous velocity is the velocity of a particle at a specific moment in time. Average velocity gives an overall picture of the particle's motion, while instantaneous velocity provides information about its motion at a specific point in time.

## How does acceleration affect particle velocity?

Acceleration is the rate of change of velocity. When a particle experiences acceleration, its velocity changes over time. If the acceleration is in the same direction as the velocity, the particle will speed up. If the acceleration is in the opposite direction, the particle will slow down. If there is no acceleration, the particle will continue to move at a constant velocity.

## How can I calculate particle velocity from a given position-time graph?

To calculate particle velocity from a position-time graph, you can find the slope of a secant line between two points on the graph. This will give you the average velocity over that time interval. To find the instantaneous velocity at a specific moment, you can find the slope of a tangent line at that point on the graph.

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