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PLEASE HELP ME suppose you ride a bicycle around the ,returning to your starting point. At the end of your trip is your average speed greater than, less than, or equal to the magnitude of your average velocity? Explain

Respuesta :

MathPhys

Answer:

Greater than

Explanation:

Distance is the length of the path traveled.  Displacement is the difference between the final position and initial position.

Speed = distance / time

Velocity = displacement / time

Since you return to your starting point, your displacement is 0 m, so your average velocity is 0 m/s.  Therefore, your average speed is greater than your average velocity.

fichoh

Speed ls a function of distance and time, while velocity is a function of displacement and time. Since the rider returned to his starting point, then the magnitude of the average speed is greater than the magnitude of the average velocity

  • Average speed = [tex]\frac{distance}{time} [/tex]

  • Average velocity = [tex]\frac{displacement}{time} [/tex]

The total distance traveled will be sum of the 'to' and 'fro' distance.

The total displacement will be 0 (Final position - Initial position) ; since the rider returned to his starting position.

This means the average velocity will be 0 while the average speed will have a value greater than 0.

Therefore, the average speed will be greater.

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