Velocity vs time graph of runner

[npena29]

velocity vs time graph!

hola everyone! i have posted this before but not sure it went through...i would appeciated it is someone would help me with it.

The world record in the 100m dash set by usain "lighting" bolt at the 2009 world championships, is 9.58s. draw a velocity-vs-time graph of his run, taking into account the following information:
1. acceleration is not constant.
2. the maximum instantaneous speed that a human being can reach is 13.8m/s
3. a 100 m runner accelerates up to his peak speed early in the race, sustains it as long as he can, but is already slowing down in the late stages of the race.

the graph should be drawn carefully enough that its possible to see that the area under the curve represents the 100m distance run.


the graph i have, i labeled the x-axis as m/s for the 13.8...and i converted Bolts speed 9.58 as 10.4 m/s² and the yaxis as meters for the 100m dash. ok from there, what's the next step? how would i graph it??

 

[collinsmark]

Hello9 npena29,

Welcome to Physics Forums!

The world record in the 100m dash set by usain "lighting" bolt at the 2009 world championships, is 9.58s. draw a velocity-vs-time graph of his run, taking into account the following information:

[...snip...]

the graph i have, i labeled the x-axis as m/s for the 13.8...and i converted Bolts speed 9.58 as 10.4 m/s² and the yaxis as meters for the 100m dash. ok from there, what's the next step? how would i graph it??

Make the x-axis time (units of seconds [Hint: 0 sec represents the start of the race. The race ends in 9.58 sec])

Make the y-axis velocity (units m/s).

[Edit: Whenever you are asked to make a plot of "this" vs. "that", it is implied "this" is the y-axis and "that" is the x-axis.]

 

[npena29]

ok...so what happens to the 100m dash?

 

[collinsmark]

ok...so what happens to the 100m dash?

It's the area under the curve!

It sounds like you get to create the curve on your own. If you want to keep it simple, you might wish to combine a right triangle and a trapezoid (plus maybe a rectangle too), but I'm guessing you're free to be creative. Choose the dimensions carefully such that the maximum height is what you desire (Hint: that 13.8m/s should fit right in here somewhere), and the overall area is 100 m.

 

[npena29]

i have attached the graph..does it look something like that?

 

[collinsmark]

i have attached the graph..does it look something like that?

The x-axis looks pretty good.

The y-axis needs some work. The way you have drawn it, starting from zero, the first tick is 4 ms, implying that there is 4 m/s per tick. But the next 4 ticks go from 4 to 12 m/s (a difference of 8 m/s); that's only 2 m/s per tick. Then the last two ticks only span 1.8 m/s. Redo the y-axis, making all the ticks evenly spaced.

To the right you have the equation

\frac{100 \ \mathrm{m}}{9.58 \ \mathrm{sec}} = 10.44 \ \mathrm{m/s},

which is good to know. That is Usain "Lighting" Bolt's average speed. So when it comes time to make your plot, you know that his instantaneous speed will sometimes be above that number, sometimes below.

Then you have 9.58 s = 10.44 m/s. I have no idea what that means.

Eventually you will need to make the actual plot. Like I mentioned before your life will be a lot easier if you stick to using straight lines. To start, you'll need to make a straight line from 0 up to some speed less than or equal to 13.8 m/s (but you should probably try to stay a little less than that). From there you need a couple more lines, but you can figure these out.

Like the problem statement said, take care to ensure the total area under the curve represents the 100 m. So you should probably do a few calculations first. Make sure you know the area formulas for a triangle, a square, and a trapezoid. Choose the dimensions of these shapes carefully, so their total area is 100 m.

(If you're wondering why area is in units of meters here, look at the axes of the graph. One dimension, length, is seconds . The other dimension, height, is meters per second [m/s], so the units of the area, length times height, [m/s] = [m].)

 

[npena29]

i got it! thank you very much