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"A runner runs at 4 m/s on a circular monitor constantly. A and also B space opposite clues on the track, therefore A is the beginning point and also B is halfway about the track. Calculation his adjust in velocity indigenous A come B, if any."

I recognize that velocity is a vector, and also therefore has a direction. Logically his adjust to me is -8 m/s. As -4 - 4 = -8, however is the adjust 8 or -8? Or is the 0 because 4 - 4 = 0?

Help is appreciated.


If you specify your axes in such a method that the runner has actually a rate of 4m/s in the y direction in allude A, the the velocity in point A is:

$V_A = 4 , extm/s , haty + 0 , extm/s , hatx$

Your runner must have a rate of -4m/s in the y direction in point B, you gain the velocity:

$V_B = -4 , extm/s , haty + 0 , extm/s , hatx$

The speed in the x direction is unchanged.

Regardless of an option of axis the equation need to be: $V_A + change = V_B = -V_A$, therefore the adjust in this situation must constantly be equal to $-2 cdot V_A$

You initial guess: v of -8m/s, v regard to the magnitude of the velocity change, is correct if you location your early vector together with one of the axes of her coordinate mechanism of choice.


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edited Nov 15 "15 in ~ 16:44
answered Nov 15 "15 at 16:36

A. NielsenA. Nielsen
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Change that velocity is definitely -8m/s, follow to vector summation.Magnitude of change of velocity is 8m/s. "-" sign suggests that the change is taking in the various other direction,i.e. In the direction opposite to the initial direction. 4-4=0 is the readjust in speed , no velocity

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answer Nov 15 "15 at 16:37

Syed JaffriSyed Jaffri
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