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?
 
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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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