What is the average speed of a particle with given position and time data?

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SUMMARY

The average speed of a particle with the position function x = (-5.06 m/s)t + (3.05 m/s²)t² from t = 0 to t = 1.20 s can be calculated using integration techniques. The correct approach involves differentiating the position function to find the velocity, then integrating the absolute value of the velocity over the specified time interval. The final average speed is determined by dividing the total distance traveled by the total time, ensuring to account for any changes in direction within the interval.

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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!
 
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you can differentiate to get the velocity then take the integral of its absolute value then divide it by time
 
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.
 
did you make sure to separate it into two integral when you integrated the absolute value one negative and one positive?
 
(-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.
 
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
 
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.
 
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
 
I came up with 2.51m/s but it is still showing as incorrect. Is this the solution that you arrived at?
 
  • #10
did you divide it by the time?
 
  • #11
Thank you! You just saved my grade!
 

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