Calculate the average velocity of the car

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Homework Help Overview

The problem involves calculating the average velocity of a car described by a position function x(t) = bt² - ct³, with specific values for b and c. The time interval for the calculation is from t = 0 to t = 10.0 s.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of average velocity and the correct interpretation of the formula vav = (V2 - V1) / (T2 - T1). There are questions about the units of velocity and how they relate to the average velocity calculation. Some participants express confusion about the distinction between average velocity and average acceleration.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions, and clarifying definitions. Some have attempted calculations but noted discrepancies in their results. The discussion is exploring the correct application of the average velocity definition.

Contextual Notes

There is an emphasis on understanding that the velocity function is not linear with respect to time, which affects the calculation of average velocity. Participants are also considering the implications of the given position function on their calculations.

David112234
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Homework Statement


[/B]
A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)=bt2−ct3, where b = 3.00 m/s2 and c = 0.130 m/s3 .

Calculate the average velocity of the car for the time interval t= 0 to t= 10.0 s

Homework Equations


[/B]
vav = V2-V1/T2-T1

The Attempt at a Solution


[/B]
x(t) = 3.00 t^2 - .130 t^3
differentiation \/
v(t) = 6.00t - .39 t^2
a(t) = 6.00 - .78t

V(0) = 6.00 (0) - .39 (0)
v(10) = 6.00 (10) - .39 (10)^2
= 60.0 - 39.
= 21

21-0/ 10-0
21/10
2.1

it tells me its wrong, why?
 
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David112234 said:
vav = V2-V1/T2-T1
What are the units of (V2-V1)/(T2-T1) or of velocity/time?

What are the units of distance divided by time?

What are the units of speed or velocity?
 
The average velocity is not the same as the change in velocity divided by time (because that is the average acceleration).
 
BvU said:
The average velocity is not the same as the change in velocity divided by time (because that is the average acceleration).
oh right, I forgot, so I tried dividing it my 2 instead of the time interval and i get 10.5, not the answer either
 
That's because the velocity doesn't depend linearly on time: ##\Delta v\over 2 ## doesn't work then.
So all you have to go by in this exercise is the definition of average velocity: $$<\vec v> \equiv {\Delta \vec x \over \Delta t}$$ and with the given information it turns out to be a fairly easy exercise :smile:

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