• Support PF! Buy your school textbooks, materials and every day products Here!

Olympic class runner

  • Thread starter grimsdale
  • Start date
1
0
1. Careful measurements have been made of Olympic sprinters in the 100 meter dash. A simple but reasonably accurate model is that a sprinter accelerates at 3.8 m/s2 for 3.26 s, then runs at constant velocity to the finish line.
(a) What is the race time for a sprinter who follows this model?

(b) A sprinter could run a faster race by accelerating faster at the beginning, thus reaching top speed sooner. If a sprinter's top speed is the same as in part a, what acceleration would he need to run the 100 meter dash in 9.61 s?

(c) By what percent did the sprinter need to increase his acceleration in order to decrease his time by 1%?




2. Homework Equations
Vf=Vi+at
Sf=Si +Vi*t+.5*a*t^2




3. The Attempt at a Solution

ok i am able to get part (a):

i started by finding the velocity for the rest fo teh race after the acceleration:
Vf=0+((3.8)(3.26))=12.388 m/s

then i found how far he traveled during the acceleration:
Sf=0+0+.5((3.8)(3.26))=20.19m

next i found how much time it took to complete the race from t=3.26s:
100=20.19+(12.388)t+.5(0)t^2
100-20.19=12.388t
t=6.44s

finally:
t(total)= 3.26+6.44= 9.7s

now i have been looking at the last two parts and i have no idea where to start.
 

Answers and Replies

587
0
What have you tried so far for parts b and c ?
To start you off on b, look at the given information. You have total distance, total time and maximum velocity. The only factors you do not know directly are the acceleration, the acceleration phase time and the constant velocity time.
Thus you have 3 eqns and 3 unknowns, I think you know what to do now :)
 
161
0
Looking at your final equation you have:

100 = 20.19 + 12.388t + .5(0)t^2

In this new situation the only thing we are changing is the 20.19 part because we don't know how far he goes in the first leg, so that part can become X lets say

100 = x + 12.388t

Now we just gotta think, what exactly is x. Well look how you calculated it previously

Sf=0+0+.5((3.8)(3.26)^2)=20.19m

Since we don't know the 3.8 or 3.26 we can just fill them as A and [itex]t_1[/itex] respectively ([itex]t_1[/itex] because it's not the same time as the time variable thats already in the equation).

[tex] 100 = (.5)(a)(t_1)^2 + 12.388t [/tex]

That is an equation with 3 unknowns. The rest is is just using known information to eliminate variables. See if you can solve it from there
 
Last edited:
92
0
Looking at your final equation you have:

100 = 20.19 + 12.388t + .5(0)t^2

In this new situation the only thing we are changing is the 20.19 part because we don't know how far he goes in the first leg, so that part can become X lets say

100 = x + 12.388t

Now we just gotta think, what exactly is x. Well look how you calculated it previously

Sf=0+0+.5((3.8)(3.26)^2)=20.19m

Since we don't know the 3.8 or 3.26 we can just fill them as A and [itex]t_1[/itex] respectively ([itex]t_1[/itex] because it's not the same time as the time variable thats already in the equation).

[tex] 100 = (.5)(a)(t_1)^2 + 12.388t [/tex]

That is an equation with 3 unknowns. The rest is is just using known information to eliminate variables. See if you can solve it from there
Can any1 explain this a little more clearly perhaps, I've been looking at this for atleast an hour and i'm still stuck....
 
92
0
anyone? im still a little stuck
 
92
0
no help?
 

Related Threads for: Olympic class runner

  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
17K
  • Last Post
Replies
1
Views
2K
Replies
9
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
10
Views
3K
Top