Question

A horse trots away from its trainer in a straight line, moving away 8m away in 8.0s it then turns abruptly gallops halfway back in 2.0’s what is the magnitude of the horses average velocity for the entire trip

Answer 1

Answer:

v = 0.4 m/s

Explanation:

• By definition, the average velocity is the rate of change of the position with respect to time, as follows:

       $v_{avg} = \frac{x_{f}-x_{o}}{t_{f} -t_{o}} (1)$

• Choosing arbitrarily x₀ = 0 and t₀ = 0, (1) reduces to:

       $v_{avg} = \frac{x_{f}}{t_{f} }} (2)$

• During the first part of the trip, the horse moves 8m away from its trainer, and during the the second part, gallops halfway back, which means that it finishes 4m away from its trainer.
• Since total time is 8.0 s + 2.0 s = 10s, replacing xf = 4m and tif = 10,0 s in (2) we get:

       $v_{avg} =\frac{x_{f}}{t_{f} } = \frac{4.0m}{10s} = 0.4 m/s (3)$