Velecity of a galloping horse

[chinnie15]

Hello,

I am new to physics and am having trouble with this homework problem. I thought I understood it, but it told me I was wrong and I'm not quite sure why. Here is the motion diagram I was given in the homework for the problem:

The question is: What is the horse's velocity during the first ten seconds of its gallop?


Seems easy enough?

I know v=Δx/Δt, and Δx= x_f-x_i, and Δt= t_f-t_i

So, from that I am figuring that the x on the diagram starts at 700m. At 10 seconds, the displacement would be 500-700= -200. The time would be 10 seconds (since it's going from 0s to 10s). Then, the velocity would equal -20m/s. But, the program is telling me that I am incorrect. Am I just reading/interpreting something wrong here? I think I understand, generally, what I'm supposed to be doing?

Thank you so much for any help on this!

 

[HallsofIvy]

You do not have anything labeled "0s" but you have that rightmost dot that is not labeled. Is that "0s"? if so, that corresponds to 600 m, not 700. (700 m is not even on your graph.)

 

[chinnie15]

That is the graph that is given in the problem. I didn't create it. It says this in the problem, above the graph:
The figure shows the motion diagram for a horse galloping in one direction along a straight path. Not every dot is labeled, but the dots are at equally spaced instants of time.

I just assumed that the 'x' (which I assumed was 700m) was where the galloping started? Or is the first dot where the galloping starts? I'm so confused...

 

[chinnie15]

Shoot... just figured it out. That first dot IS the starting place (on 600m), not 'x'. So, it was -10, not -20. Got it now.

Thanks!

 

[asteriatic]

Hello,

It seems like you have the right idea in terms of using the equation v=Δx/Δt to calculate the velocity of the galloping horse. However, there may be some confusion in how you are interpreting the motion diagram.

First, it's important to note that the motion diagram is showing the displacement of the horse, not the position. This means that the starting point (xi) is actually 0m, not 700m. The displacement at 10 seconds would then be 500m, not -200m.

Secondly, the time interval (Δt) should be the change in time between the two points on the motion diagram, not the total time. In this case, the time interval would be 5 seconds (from 0s to 5s) since the horse is moving at a constant velocity during that time.

Using these values, the velocity would then be calculated as v=Δx/Δt = 500m/5s = 100m/s.

I hope this helps clarify the problem for you. It's important to pay close attention to the details and units when working with equations and diagrams in physics. Keep up the good work and don't hesitate to ask for help when needed!