Simple question about velocity and acceleration

1. Aug 27, 2005

h_k331

I'm writing a lab report and came to the following conclusion, I was hoping someone might be able to verify it for me.

The average velocity during the time interval [A,B] will be equal to the instantaneous velocity at the time (A+B)/2 if the acceleration during the time interval [A,B] is a constant.

Thanks,
hk

2. Aug 27, 2005

daveed

yeah, thats true

3. Aug 27, 2005

Andrew Mason

The graph of velocity - time will have a slope a= v/t that is constant. In other words, it is a straight line with slope a.

The definition of average speed is: $$v_{avg} = \frac{distance_{total}}{time_{total}}$$

What is the total distance covered (d = vt) in the time interval [A,B]? What is the total time?

[Hint: think of distance as the area under the graph]

AM

4. Aug 28, 2005

h_k331

Thanks Dave.

hk

5. Aug 28, 2005

h_k331

The total distance covered would be the integral of velocity, and the total time would be d/v, right?

hk

6. Aug 28, 2005

Andrew Mason

The time would be B-A.

$$t=\frac{distance_{total}}{v_{avg}}= B-A$$

$$d = \int_A^B vdt = v_A(B-A) + \frac{1}{2}(v_B - v_A)(B-A)$$

$$d/t = v_{avg} = d/(B-A) = v_A + \frac{1}{2}(v_B - v_A) = v_A + \frac{1}{2}(at)$$
$$v_{avg} = v_A + \frac{1}{2}(a(B-A)) = v_A + a\frac{(B-A)}{2}$$

which is the speed at time (B-A)/2.

AM

7. Aug 29, 2005

h_k331

Thanks Andrew.

hk

8. Aug 29, 2005

lightgrav

d/v certainly has UNITS of time,
but you really NEED to specify what distance you mean
(midpoint? location at mid-time? total distance?)
and what velocity you mean
(slowest=v_A ? fastest=v_B ? average? v at mid-point?)

time-average of velocity = total displacement/total duration