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Finding Time With Distance And Acceleration?

  • Thread starter fdsa1234
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  • #1
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Hey guys, just signed up because I haven't been able to figure this problem out. As part of a question, I have to find the time it takes for a train to decelerate from a velocity of 20 m/s at a rate of 2.4 m/s^2 over 83.3m, but I haven't been able to figure it out.

I tried multiple equations but haven't been able to get it.

Original velocity is 20 m/s, d = 83.3m, a = -2.4 m/s^2, final velocity = 0

Help?
 

Answers and Replies

  • #2
rock.freak667
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What equation do you know involves initial velocity, final velocity, time and acceleration?

(Hint: Think about what acceleration is defined as)
 
  • #3
gneill
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Hi fdsa1234. Welcome to Physics Forums.

Can you show us some of the equations you've tried, and perhaps demonstrate what you tried to do with them?
 
  • #4
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I tried using d = V1(t) + ½ at^2, not sure if I have an equation that uses V1, V2, t, and a
 
  • #5
gneill
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Are you sure that the original question specified that the final velocity is zero?

The reason I ask is that it seems to me that the parameters are over-specified if both distance criteria and final velocity criteria are specified. In fact, the two criteria may be incompatible with a "real" solution.
 
  • #6
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Well, it's a multi-part question, maybe I screwed up as we just started doing physics and I'm not that great at it... The train accelerates in part 1, then goes at a constant rate for part 2, and then in part 3 decelerates to stop at the station. At the end of part 2/beginning of part 3, velocity is 20 m/s, and then it decelerates at 2.4 m/s^2 over 83.3m (distance for part 3 was found earlier in the question).
 
  • #7
gneill
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Perhaps it would be prudent to post the entire question, along with the results that you've obtained to the various parts so far. I checked the 83.3m distance and zero velocity criteria against the other given values, and there is a (small) discrepancy that could be problematic.
 
  • #8
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83.3m was actually 83.33333...m, repeating, if that makes a difference.
 
  • #9
gneill
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83.3m was actually 83.33333...m, repeating, if that makes a difference.
Yes it does! :smile: Worrisome discrepancy abated!

It pays to carry extra precision through the workings of a problem, and then round to the required significant figures for the results only at the end.

So, where were we. Ah yes. You mentioned the equation d = V1(t) + ½ at2. That equation should give you what you want if you plug in the known values and solve for t. Note that, being quadratic in t, it will present you with two solutions that you will have to sort out.

As an alternative, a much simpler equation will do for the case where you're going from a given velocity down to zero velocity with a constant acceleration. What is the equation for velocity after time t given an initial velocity and acceleration?
 
  • #10
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Eh... V2 = V1 + at?
 
  • #11
gneill
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That looks promising. Now, if V1 is your 20 m/s, and V2 is zero...
 
  • #12
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Yeah, I tried that equation after I posted and got 8.33... s, which I believe must be correct.
 
  • #13
gneill
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It looks good. You might want to confirm that the distance covered in that time while the train is decelerating matches the given value. d = vot + (1/2)at2.
 
  • #14
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I see. Thanks a lot for your help, guys.
 

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