# Calculate the average velocity of the car

1. Nov 14, 2015

### David112234

1. The problem statement, all variables and given/known data

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

2. Relevant equations

vav = V2-V1/T2-T1

3. The attempt at a solution

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?

2. Nov 14, 2015

### Staff: Mentor

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?

3. Nov 14, 2015

### BvU

The average velocity is not the same as the change in velocity divided by time (because that is the average acceleration).

4. Nov 14, 2015

### David112234

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

5. Nov 15, 2015

### BvU

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

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