- #1

h_k331

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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

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- Thread starter h_k331
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- #1

h_k331

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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

daveed

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yeah, thats true

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Andrew Mason

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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.h_k331 said: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 definition of average speed is: [tex]v_{avg} = \frac{distance_{total}}{time_{total}}[/tex]

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

h_k331

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daveed said:yeah, thats true

Thanks Dave.

hk

- #5

h_k331

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Andrew Mason said: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: [tex]v_{avg} = \frac{distance_{total}}{time_{total}}[/tex]

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

- #6

Andrew Mason

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The time would be B-A.h_k331 said:The total distance covered would be the integral of velocity, and the total time would be d/v, right?

[tex]t=\frac{distance_{total}}{v_{avg}}= B-A[/tex]

[tex]d = \int_A^B vdt = v_A(B-A) + \frac{1}{2}(v_B - v_A)(B-A)[/tex]

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

[tex]v_{avg} = v_A + \frac{1}{2}(a(B-A)) = v_A + a\frac{(B-A)}{2}[/tex]

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

AM

- #7

h_k331

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Thanks Andrew.

hk

hk

- #8

lightgrav

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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

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