What is the acceleration of the runner?

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To determine acceleration from a curved line on a distance-time graph, one approach is to find the speed by calculating the slope of the tangent at various points. This can be visualized by creating a speed-time graph from the distance-time graph, where the slope of the speed graph represents acceleration. Alternatively, if the radius of curvature is known, centripetal acceleration concepts can be applied by modeling the scenario as a rocket in a constant horizontal wind. A method involving a mirror can also be used to accurately find tangents and thus derive the necessary slopes. Ultimately, calculating the second derivative of the distance function at multiple points provides a clearer picture of acceleration.
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acceleration from a curved line on a distance-time graph

Homework Statement


how do i work out the acceleration from a curved line on a distance-time graph?

Homework Equations




The Attempt at a Solution

 
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welcome to pf!

hi bee-gee! welcome to pf! :smile:
bee-gee said:
how do i work out the acceleration from a curved line on a distance-time graph?

you obviously know how to find the speed (from the slope of [STRIKE]area under[/STRIKE] the graph)

i suppose you could use that to draw a speed-time graph, and then measure the slope of [STRIKE]area under[/STRIKE] that

alternatively, if you can measure the radius of curvature, r, and if you know about centripetal acceleration, then you can do it by imagining that the graph represents a rocket rising vertically …

now put the rocket in a constant horizontal wind, so that sideways distance is proportional to time: if you get the constant right, the vertical-horizontal distance-distance graph will look exactly the same as the original distance-time graph, and you know the total acceleration is purely vertical :wink:
 
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tiny-tim said:
hi bee-gee! welcome to pf! :smile:


you obviously know how to find the speed (from the area under the graph)


Isn't speed the slope of a distance / time graph?:confused:
 
oops!

oops! :redface:

i typed too quickly! :biggrin:

thanks, i'll edit it :smile:
 
Yes. I'm sure tiny-tim just mis-spoke. sjb-2812, a cute method I was taught, in high school, for finding tangents, and so derivatives, from a graph, used a small mirror. Hold the mirror on the graph at the point at which you want the tangent. Rotate the mirror about the vertical axis until the graph seems to go smoothly into its reflection in the mirror. Use the mirror as a straight edge to draw a line perpendicular to the curve. Now turn the mirror to draw a line perpendicular to that line and repeat the process to draw a line perpendicular to the original perpendicular and so tangent to the curve. From the values on the graph of the endpoints of that tangent line, you can find the slope of the tangent line and so the derivative of the function.

To find the acceleration, the second derivative of distance, you would need to do that in a number of different places to get a rough graph of "velocity versus time" and then repeat the process with that graph.

The simplest thing to do is to approximate the speed, at a number of points using the difference quotient of a number of near by points. Then, again, repeat the process to find the second derivative.
 
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