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Kinematics - average velocity

  • Thread starter Hassin
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  • #1
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Hello, I have problem with this example. I don't speak English very well, so please forgive me if something isn't quite enough defined - as it should be in English Physics terminology.

The train moves from the station with an acceleration a. After reaching velocity v moves farther with constant velocity, next train moves with deceleration and, at last to dwell on the following station. What is the average velocity of the train v(avr) on the rout in length s?

I know that the answer should be v(avr) = v/(1+(v^2)/as). But I don't know how to get to it. Could somebody tell me step by step what should I do?
 

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  • #2
haruspex
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You didn't specify the deceleration rate, but from the answer it looks as though that is also a (or rather, -a). Suppose the acceleration lasted time t.
What is the relationship between a, v and t?
How far did it travel while accelerating?
How long did it spend decelerating?
How far did it travel while decelerating?
So how far did it travel at speed v?
So how long did it spend at speed v?
So how much time did the whole journey take?
 
  • #3
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Let the magnitude of acceleration and deceleration is equal to a

Let d the distance travelled both acceleration and deceleration

d= v2/a
s=d+Tv, Time for constant velocity

T=s/v - d/v= s/v-v/a

Vav=s/(T+2v/a)

Vav=s/(s/v-v/a+2v/a)

Vav=s/(s/v+v/a)

Vav=s/((sa +v2)/va)

Vav=sav/(sa +v2)

Vav=v/(1 +v2/sa)
 
Last edited:
  • #4
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Thank you very much, I didn't know this formula without time d=(v^2)/2a.
Very helpful :)
 
  • #5
1,065
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By drawing a velocity vs. time graph, you can easily see the distance covered every segment.
 

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