Distance/speed problem

  • Thread starter wadesweatt
  • Start date
  • #1
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A horse canters away from its trainer in a straight line, moving 140 m away in 14.0 s. It then turns abruptly and gallops halfway back in 4.3 s.

(a) Calculate its average speed. (m/s)

(b) Calculate its average velocity for the entire trip, using "away from the trainer" as the positive direction. (m/s)


To start, I have drawn a graph to represent this situation, with time on the x-axis and distance on the y-axis. 1) I don't know how average speed and average velocity are different (since I thought they were both how fast something is going, or the slope of the graph), and 2) I don't know how to find the derivative of the graph since it is not linear and I am given no additional information about the equation of the graph.

Maybe I am thinking way too far into this. I am new to physics, but I've had lots of math.

Thanks in advance for the help.

Wade
 

Answers and Replies

  • #2
79
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Well if you know anything about vectors and scalars, then the difference should be easy to realize that speed is a scalar value and velocity is a vector value.

average speed = total distance covered / time

average velocity = total displacement / time
 
  • #3
33
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Thanks for the explanation. Yep, I've never had any college or high school physics before so I had never heard about that difference.
 

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