Average speed of a particle

  • Thread starter Casimi
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  • #1
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Homework Statement


The position of a particle as a function of time is given by x = (-5.06 m/s)t + (3.05 m/s2)t2. Calculate the average velocity of the particle from t = 0 to t = 1.20 s.
-1.40 m/s (this is correct, from my calculations)

2nd part: Calculate the average speed from t = 0 to t = 1.20 s


The Attempt at a Solution


I have tried this for hours and cannot arrive at the correct answer for the second part. I tried taking the total distance and dividing by the total time. Could someone please point me in the right direction to find average speed? I have even tried differentiation.

Any assistance would be appreciated. Thanks!
 

Answers and Replies

  • #2
326
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you can differentiate to get the velocity then take the integral of its absolute value then divide it by time
 
  • #3
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I tried that but I am still not arriving at the correct answer. I arrive at a number far larger than it should be.

I cannot figure out what I am doing wrong.
 
  • #4
326
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did you make sure to separate it into two integral when you integrated the absolute value one negative and one positive?
 
  • #5
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(-5.06+6.10t) - that is my answer after differentiating. Integrating the absolute value, I have 5.06t+3.05t^2 with integral limits of 0 to 1.20. When I put the values in the integral, I get 9.456 and when divided by 1.2, I get 7.88 which is not the correct answer.
 
  • #6
326
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I am saying as integral the absolute value
you must know if when v = 0 lies in your interval
-5.06+6.10t=0
t=5.06/6.10 this lies in the interval so you make the integral negative here and positive from it to 1.2
 
  • #7
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Yes, I tried that and am still not arriving at the right answer. 5.06/6.10 will give .82. If I integrate from -5.06t+3.05t^2 [0 to .82] + 5.06t+3.05t^2 [.82 to 1.2] I am still going to get :
5.06t +3.05t^2 [0 to 1.2]???? Please help me out.
 
  • #8
326
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what you do is get the integral from 0 to .82 (5.06-6.1t)dt
+ the integral from .82 to 1.2 (-5.06 +6.1t)dt
 
  • #9
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I came up with 2.51m/s but it is still showing as incorrect. Is this the solution that you arrived at?
 
  • #10
326
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did you divide it by the time?
 
  • #11
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Thank you! You just saved my grade!
 

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