Acceleration vs. Time

  • #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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Answers and Replies

  • #2
teclo
117
0
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
stunner5000pt
1,449
2
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
onetroubledguy
6
0
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.
 

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