# Sprinter Problem

b_vishwajit used Ask the Experts™
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The 100m dash can be run by the best sprinters in 10.0s. A 62kg sprinter accelerates uniformly for the first 45m to reach top speed, which he maintains for the remaining 55m. (a) What is the average horizontal component of force exerted on his feet by the ground during accelertion? (b) What is the speed of the sprinter over the last 55m of the race(i.e., his top speed.).

Step by step solution required. This is a solved problem. I have the answers with me. Just want to see different methods. Have phun!!!!!!!!!!1
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Commented:
b_vishwajit,
If t1 is the time for the 1st 45m and t2 the time for the 2nd 55m and a is the uniform accelaration then

(a*t1^2)/2 = 45        (1)      (by newtons equations)

a*t1*t2    = 55        (2)      (a*t1 = final velocity)

t1+t2    = 10        (3)      (total time = 10 secs)

dividing eqn 2 by eqn 1

a*t1/(2*t2)  = 9/11

substitute t2 = 10 - t1 gives

t1/(2*(10 - t1)  = 9/11

solving for t1 gives t1 = 6.20689s

giving (using eqn 3) t2 = 3.7931s

using 2 gives  a = 55/(6.20689*3.7931) = 2.33611m/s

since f=m*a then average horizontal force = 2.33611*62 = 144.8387 NMs(-1)

cheers
GfW

Commented:
a*t1/(2*t2)  = 9/11 should read  t1/(2*t2)  = 9/11.

knowing me there are arithmetic mistakes but that is the basic idea

Commented:
ps top speed is a*t1 = 6.20689*2.33611 = 14.499m/s (ie really fast)
Commented:
vf = final speed in meters/second
t1 = duration of constant acceleration phase in seconds
t2 = duration of constant speed phase in seconds
d1 = distance covered during constant acceleration phase = 45 meters
d2 = distance covered during constant speed phase = 55 meters

Average speed * time = distance covered during constant acceleration:
(vf/2)*t1 = d1

Constant speed * time = distance covered during constant speed:
vf*t2 = d2

Total time is 10 seconds:
t1 + t2 = 10

From first equation get t1 = 2*d1/vf and plug into last equation to get t2 in terms of vf:
t2 = 10 - 2*d1/vf

Use this t2 in second equation to get
vf*10 -2*d1 = d2

Then solve for vf:
vf = (d2 + 2*d1)/10 = 14.5 m/s

and
t1 = 2*d1/vf = 6.21 s

so
a = vf/t1 = 2.34 m/s/s

Force = mass * acceleration:
F = 62 kg * 2.34 m/s/s = 144.8 N
Commented:
More or less the same approach as above (there really aren't too many ways to do such a problem):

0                    45                      100
|        a->        |                        |
|        t1          |        t2             |
v

45 = (1/2) a t1^2  ==>  a t1^2 = 90                (1)

v = a t1                                                (2)

55 = v t2  ==>  v = 55 / t2                        (3)

So,

a t1 = 55 / t2  ==>  a t1 t2 = 55                        (4)

t1 + t2 = 10  ==> t1 = 10 - t2                        (5)

Plugging (5) into (4),

a t1 (10 - t1) = 55                                (6)

From (1), we have

a = 90 / t1^2                                        (7)

Plugging (7) into (6), we have

t1 (10 - t1)                      55
---------------        =        ------                        (8)
t1^2                         90

10 - t1                              11
----------                =        -----                        (9)
t1                                  18

10 - t1 = (11/18) t1  ==>  10 = (1 + 11/18) t1        (10)

This is just 10 = (29/18) t1                        (11)

So,

t1 = 180 / 29 (seconds)                                (12)

Now, from (7),

90 * 29^2                29^2
a = 90 / (180/29)^2 =         ------------------ =        --------
180 * 180                 360

62 * 29^2                   31 * 29^2
So, F = m a =        ---------------        =        -------------- = 144.8389 N
360                              180

29^2            180               29
v = a t1 =        --------   x        ------   =        ----    =        14.5 m/s
360                29                2
Commented:
Simpler approach using areas in velocity graph

v|
|
|
|_________ vf
|      /|       |
|45 / |       |
|    /  | 55  |
|   /   |       |
|  /45|       |
| /     |       |
|/___|___ |____ t
0       t1      t2

45 + 45 + 55 = 145 = 10 * vf    So  vf = 14.5 m/S

45 = 1/2 * vf * t1    So t1 = 90/14.5 = 6.21 S

F = m * a = m * vf / t1 = 62 * 14.5 / 6.21 = 144.8 N

Commented:
wytcom: Hmm, looks the same as your first answer really but with less detail. :-)

Commented:
I plead "No Contest"

But using the area of the full rectangle is what is new here.

Commented:
I am just ragging you. Velocity diagrams can give insight into problem.  :-)

Commented:
His feet don't always touch the ground.  If you are averaging the force for the whole time, this is fine.  If you want to find the actual average force, you would need to know how long his feet are in contact with the ground.

Commented:
" If you want to find the actual average force, you would need to know how long his feet are in contact with the ground."

No, if you knew that you could calculate the IMPULSIVE force. The AVERAGE force does not need this information since it is the integral of all impulsive forces divided by the time in which the forces acted.

Commented:
BR, you just said the same thing I was:

The AVERAGE force does not need this information since it is the integral of all impulsive forces <emphasis>divided by the time in which the forces acted<end emphasis>.

When you're mid stride (ie, not touching the ground) you are not applying any force.  So in order to find the average applied force, you need to know how long each force was applied (duh! :) ), otherwise you average out with a lot of zero force time.
Commented:
t1 = time for 45m
a = accleration
then we have

0.5 *( a*t1^2) = 45
=> a*t1^2 = 90                                           -(a)
and a*t1*(10 - t1) = 55
=> 10*a*t1 - a*t1^2 = 55                            -(b)
putting a in b

10*a*t1 = 55 + 90 = 145
=> a*t1 = 14.5
=> 14.5 * t1 = 90
=> t1 = 6.206
=> a = 90 / (6.206)^2 = 2.336

=> velocity for 55m = 2.336*6.206 = 14.5m/s
force = mass * acc

=> force = 62kg* 2.336 = 144.832N

these are your answers, now awards some points to the correct answers and approaches.
tata

Commented:
hey guys talking about acceleration he has clearly stated in the question that the sprinter maintains a constant accleration for the first 45m hence there is a constant net force being applied for the 45m hence it is clear that at every instance in the 45m atleast one leg is in contact with the ground.

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