Finding runner acceleration

[osker246]

Homework Statement

A simple model for a person running the 100 m dash is to assume the sprinter runs with a constant acceleration until reaching top speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of 11.2 m/s in 2.14 s, what will be his total time?

Homework Equations

Xf = Xi + ((Vx)i)(Delta T) + .5(Ax)(Delta T)^2

((Vx)f)^2 = ((Vx)i)^2 + 2(Ax)(Delta X)

The Attempt at a Solution

Ok so I need a second opinion to know if I've done this correctly or not. I have the textbook answer but I don't know if its correct or not since I'm not getting the same answer.

I start out by finding the acceleration while the runner gets up to speed.

11.2 m/s / 2.14 = 5.23 m/s^2

Using the info given I try and find the distance covered:

11.2^2 = 0^2 + 2(5.23)(Delta X)

Delta X = 125.44/10.46 = 11.99 m

I then find the time taken to cover the distance:

11.9 = .5(5.23)(Delta T)^2

(Delta T)^2 = 11.9/2.615 = 4.585
Delta T = 2.14 s

OK so now...

100 m - 11.99m = 88.01m

I then find the time it takes to travel this distance.

88.01 = .5(5.23)(delta T)^2

(delta T)^2 = 33.65

delta T= 5.80 sec

so I add the two times

5.80 sec + 2.14 sec = 7.94 sec

Am I doing this correctly? My textbook is showing me the time as being 10 seconds but I don't see any other way of finding this answer. Any help is appreciated! Thanks!

 

[alphysicist]

Hi osker246,

Homework Statement

A simple model for a person running the 100 m dash is to assume the sprinter runs with a constant acceleration until reaching top speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of 11.2 m/s in 2.14 s, what will be his total time?

Homework Equations

Xf = Xi + ((Vx)i)(Delta T) + .5(Ax)(Delta T)^2

((Vx)f)^2 = ((Vx)i)^2 + 2(Ax)(Delta X)

The Attempt at a Solution

Ok so I need a second opinion to know if I've done this correctly or not. I have the textbook answer but I don't know if its correct or not since I'm not getting the same answer.

I start out by finding the acceleration while the runner gets up to speed.

11.2 m/s / 2.14 = 5.23 m/s^2

Using the info given I try and find the distance covered:

11.2^2 = 0^2 + 2(5.23)(Delta X)

Delta X = 125.44/10.46 = 11.99 m

I then find the time taken to cover the distance:

11.9 = .5(5.23)(Delta T)^2

(Delta T)^2 = 11.9/2.615 = 4.585
Delta T = 2.14 s

OK so now...

100 m - 11.99m = 88.01m

I then find the time it takes to travel this distance.

88.01 = .5(5.23)(delta T)^2

The runner is not accelerating once he reaches top speed, so I think this step is not right.

 

[osker246]

Hi osker246,



The runner is not accelerating once he reaches top speed, so I think this step is not right.

wow I cannot believe I over looked that. So then it just comes down to using T= D/R to find the rest of the problem. Thanks A lot alphysicist! I've been stumped on this problem for the past hour trying to figure out where I went wrong.

 

[alphysicist]

wow I cannot believe I over looked that. So then it just comes down to using T= D/R to find the rest of the problem. Thanks A lot alphysicist! I've been stumped on this problem for the past hour trying to figure out where I went wrong.

Yes, once you use that equation (T=D/R), the rest looks perfect. And by the way, the way you wrote your original post was great; giving the details of your work made it very easy to understand what you did.