Simple kinematics sprinter problem

  • Thread starter david90
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In summary, a sprinter reaches his max speed vmax in 2.5 sec from rest with const accel. This sprinter then maintains that speed and finishes the 100 yards in the overall time of 9.6 sec. His max speed vmax is 12 ft/sec.
  • #1
david90
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A sprinter reaches his max speed vmax in 2.5 sec from rest with const accel.

He then maintains that speed and finiishes the 100 yards in the overall time of 9.6 sec. Dettermine his max speed vmax.
 
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  • #2
i got a = 4.8 ft/s^2 thus vmax = 4.8*2.5sec = 12ft/s

seens a bit fast.
 
  • #3
If you post the equations you've worked with, and how you attacked them, rather than posting a couple of numbers, it would be easier to spot possible mistakes you may have done.
 
  • #4
david90 said:
A sprinter reaches his max speed vmax in 2.5 sec from rest with const accel.

He then maintains that speed and finiishes the 100 yards in the overall time of 9.6 sec. Dettermine his max speed vmax.

Sketch a speed time graph and calculate with the fact area = distance.
 
  • #5
sorry david, but 12 ft/sec as a max speed is a little to slow...this guy here's running 300 ft (100 yrds) in 9.6 seconds.
 
  • #6
david90, here's a way you can solve it algebraicly:

For the first 2.5 seconds, the sprinter is accelerating...thus...we can describe that part as

distance = (1/2)at^2, with t = 2.5 it would be = (1/2)a(2.5)^2

For the last 7.1 seconds, we know that the speed is both constant and max...and also that the speed = a * 2.5...since distance = speed * time (which in this case is the latter 7.1)...you get...

distance = (a * 2.5) * 7.1

Now, just add these two equations above together, and you get the total distasnce. Set this equal to 300 ft, which is the total distance, and solve for acceleration. Once you know acceleration, you can then find the max. velocity.
 
  • #7
actually, based on your answer of a = 4.8 ft/sec^2...it appears that you must have used the above method, or a similar one...but instead of having the length be equal to 300 ft (100 yards)...looks like you had it equal to 100 ft. So anyways, I'm willing to guess that you had your method correct, but just forgot to convert your units. Just make sure in the future to keep your units consistent. :-)
 

1. What is "simple kinematics sprinter problem"?

"Simple kinematics sprinter problem" is a physics problem that involves analyzing the motion of a sprinter in a straight line, using basic concepts of kinematics such as displacement, velocity, and acceleration.

2. Why is the sprinter considered a simple kinematics problem?

The sprinter is considered a simple kinematics problem because it involves motion in a straight line, without any external forces or factors that can affect the sprinter's motion, such as air resistance or varying terrain.

3. What are the key equations used to solve a simple kinematics sprinter problem?

The key equations used to solve a simple kinematics sprinter problem are the equations for displacement, velocity, and acceleration:
- Displacement (Δx) = final position (xf) - initial position (xi)
- Average velocity (Vavg) = displacement (Δx) / time (Δt)
- Acceleration (a) = change in velocity (Δv) / time (Δt)

4. How is the acceleration of a sprinter calculated in a simple kinematics problem?

The acceleration of a sprinter is calculated by dividing the change in velocity (final velocity - initial velocity) by the time it took to make that change. This can also be expressed as the slope of the velocity vs. time graph for the sprinter's motion.

5. What are some real-world applications of simple kinematics sprinter problems?

Simple kinematics sprinter problems can be used to analyze the performance of athletes in track and field events, to design training programs for sprinters, and to improve the efficiency of sprinting techniques. They can also be applied in other fields, such as analyzing the motion of vehicles on a straight road or the motion of projectiles in physics experiments.

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