Simple question about velocity and acceleration

[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

 

[daveed]

yeah, that's true

 

[Andrew Mason]

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.

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

 

[h_k331]

yeah, that's true

Thanks Dave.

hk

 

[h_k331]

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

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

hk

 

[Andrew Mason]

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

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

 

[h_k331]

Thanks Andrew.

hk

 

[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