Finding Position at a Specific Time Using Kinematic Equations

In summary, the graph shows that the particle starts from rest at the initial position 0m and has an initial velocity of 0m/s. At time t=3s, the particle's position is 4.5m. At time t=2s, the particle's position is 6.5m and the total distance travelled is 11m.
  • #1
onetroubledguy
6
0
Ok here's the graph:

http://img268.imageshack.us/img268/8762/acvt9nz.jpg [Broken]

We are given that the particle starts from rest at the initial position 0m with the initial velocity 0m/s. I need to find what the position of the particle is at t=5s.

Since acceleration is changing here, I didn't know what to do so I broke the problem up. First, I wanted to find the position at t=3s. I used the equation
x = x0 + v0t + 1/2at^2 where x0 = 0m, v0 = 0m/s, a = 1m/s^2, and t = 3s.
I came up with 4.5m.

Next, I did the same thing for the lower part; I used the same equation but with x0 = 4.5m, v0 = 3m/s, a = -2m/s^2, and t = 2s. I came up with 6.5m and added it to first distance (4.5m) to get 11m.

Obviously this is wrong. Any help you can provide would be much appreciated. Thank you.
 
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  • #2
onetroubledguy said:
Ok here's the graph:

http://img268.imageshack.us/img268/8762/acvt9nz.jpg [Broken]

We are given that the particle starts from rest at the initial position 0m with the initial velocity 0m/s. I need to find what the position of the particle is at t=5s.

Since acceleration is changing here, I didn't know what to do so I broke the problem up. First, I wanted to find the position at t=3s. I used the equation
x = x0 + v0t + 1/2at^2 where x0 = 0m, v0 = 0m/s, a = 1m/s^2, and t = 3s.
I came up with 4.5m.

Next, I did the same thing for the lower part; I used the same equation but with x0 = 4.5m, v0 = 3m/s, a = -2m/s^2, and t = 2s. I came up with 6.5m and added it to first distance (4.5m) to get 11m.

Obviously this is wrong. Any help you can provide would be much appreciated. Thank you.


wouldn't it be just the answer to your second part? you're starting at the point where the first part left off. you don't need to add that onto your answer as you've already accouned for the position up to 3 seconds with the [tex]x_0[/tex] in your second kinematic eqation.
 
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  • #3
your individual calculations are correct. However, you have to be careful of the SIGN of the acceleration in each part. What is the acceleration (as read fro mteh graph) for the first 3 seconds and then the final 2 seconds?
 
  • #4
teclo said:
wouldn't it be just the answer to your second part? you're starting at the point where the first part left off. you don't need to add that onto your answer as you've already accouned for the position up to 3 seconds with the [tex]x_0[/tex] in your second kinematic eqation.

You're right. I always do stupid things like that! Grr...thanks.
 

1. What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It can be positive (speeding up), negative (slowing down), or zero (constant velocity).

2. How is acceleration calculated?

Acceleration is calculated by dividing the change in an object's velocity by the change in time. The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between acceleration and velocity?

Velocity is a measure of an object's speed and direction, while acceleration is a measure of how quickly that velocity is changing. In other words, velocity tells us how fast an object is moving, while acceleration tells us how that speed is changing.

4. How does acceleration affect an object's motion?

Acceleration affects an object's motion by changing its velocity. If an object is accelerating, it is either speeding up or slowing down. This can result in changes in the object's position and/or direction of motion.

5. How can acceleration be represented on a graph?

Acceleration can be represented on a graph by plotting the change in an object's velocity on the y-axis and the corresponding change in time on the x-axis. The slope of the line on the graph represents the object's acceleration, with a steeper slope indicating a greater acceleration.

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