Calculate the average velocity of the car

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David112234
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Homework Statement


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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


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vav = V2-V1/T2-T1

The Attempt at a Solution


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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?
 
on Phys.org
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?
 
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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