Modeling a \(\mathbf{1 0 0} \mathbf{m}\) race. In this project we will develop
an advanced model for the motion of a sprinter during a \(100 \mathrm{~m}\)
race. We will build the model gradually, adding complications one at a time to
develop a realistic model for the race.
(a) A sprinter is accelerating along the track. Draw a free-body diagram of
the sprinter, including only horizontal forces. Try to make the length of the
vectors correspond to the relative magnitudes of the forces.
Let us assume that the sprinter is accelerated by a constant horizontal
driving force, \(F=400 \mathrm{~N}\), from the ground all the way from the start
to the \(100 \mathrm{~m}\) line (averaged over a few steps). The mass of the
sprinter is \(m=80 \mathrm{~kg}\).
(b) Find the position, \(x(t)\), of the sprinter as a function of time.
(c) Show that the sprinter uses \(t=6.3 \mathrm{~s}\) to reach the \(100
\mathrm{~m}\) line.
This is a bit fast compared with real races. However, real sprinters are
limited by air resistance. Let us introduce a model for air resistance by
assuming that the air resistance force is described by a square law:
$$
D=(1 / 2) \rho C_{D} A(v-w)^{2}
$$
where \(\rho\) is the density of air, \(A\) is the cross-sectional area of the
runner, \(C_{D}\) is the drag coefficient, \(v\) is the velocity of the runner,
and \(w\) is the velocity of the air. At sea level \(\rho=1.293 \mathrm{~kg} /
\mathrm{m}^{3}\), and for the runner we can assume \(A=0.45 \mathrm{~m}^{2}\),
and \(C_{D}=1.2\). You can initially assume that there is no wind: \(w=0
\mathrm{~m} / \mathrm{s}\).
Assume that the runner is only affected by the constant driving force, \(F\),
and the air resistance force, \(D\).
(d) Find an expression for the acceleration of the runner.
(e) Use Euler's method to find the velocity, \(v(t)\), and position, \(x(t)\) as a
function of time for the runner. The runner starts from rest at the time \(t=0
\mathrm{~s}\). Plot the position, velocity and acceleration of the runner as a
function of time. How did you decide on the time-step \(\Delta t ?\) (Your
answer should include the program used to solve the problem and the resulting
plots).
(f) Use the results to find the race time for the \(100 \mathrm{~m}\) race.(g)
Show that the (theoretical) maximum velocity of a runner driven by these
forces is:
$$
v_{T}=\sqrt{2 F /\left(\rho C_{D} A\right)} .
$$
The runner may have to run more than \(100 \mathrm{~m}\) to reach this velocity.
(We often call this maximum velocity the terminal velocity-"terminal" because
the velocity increases until it reaches the terminal velocity, where the
acceleration becomes zero). Find the numerical value of the terminal velocity
for the runner. Do you think this is realistic?
So far the model only includes a constant driving force and air resistance.
This is clearly a too simplified model to be realistic. Let us make the model
more realistic by adding a few features.
First, there is a physiological limit to how fast you can run. The driving
force from the runner should therefore decrease with velocity, so that there
is a maximum velocity at which the acceleration is zero even without air
resistance. While we do not know the detailed physiological mechanisms for
this effect, we can make a simplified force model to implement the effect by
introducing a driving force, \(F_{D}\), with two terms: a constant term, \(F\),
and a term that decreases with increasing velocity, \(F_{V}\) : \(F_{V}=-f_{V}
v\), so that the driving force is:
$$
F_{D}=F+F_{V}=F-f_{V} v .
$$
Reasonable values for the parameters are \(F=400 \mathrm{~N}\), and \(f_{v}=25.8
\mathrm{sN} / \mathrm{m}\). (These values are chosen to make the maximum
velocity reasonable - they are not based on a physiological consideration).
(h) If you assume that the runner is subject only to these two driving forces,
what is his maximum velocity? (You can ignore the drag term, \(D\), in this
calculation).
In addition, during the first few seconds the runner is crouched and
accelerating rapidly. In this phase, his cross-sectional area is smaller
because he is crouched, and the driving force exerted by the runner is larger
than later. Let us also introduce these aspects into our model.
First, let us assume that the crouched phase lasts approximately for a time,
\(t_{c}\). We do not expect this phase to end abruptly at a specific time.
Instead, we expect the driving force to decrease gradually (and the cross-
sectional area to increase gradually) as the runner is going from a crouched
to an upright running position. A common way to approximate such a change is
through an exponential function that depends on the time and the
characteristic time, \(t_{c}\). For example, by introducing an initial driving
force, \(F_{C}\) :
$$
F_{C}=f_{c} \exp \left(-\left(t / t_{c}\right)^{2}\right) .
$$
When \(t=0\), the force is equal to \(f_{c}\), but as time increases, the force
decreases rapidly. When the time has reached \(t_{c}\), the force has dropped to
\(1 / e \simeq 0.37\) of the value at \(t=0\), and after a time \(4 t_{c}\) this
contribution to the driving force has dropped to less than \(2 \%\) of its
initial value.
Notice that we do not have any experimental or theoretical reason to use this
particular form for the time dependence. We have simply chosen a convenient
form as a first approximation, and then we use this form and try to get
reasonable results with it. A better approach would be to have experimental
data on how the force varied during the first few seconds, but unfortunately
we do not know this. Making rough estimates that you can subsequently improve
by better measurements, calculations, or theory will be an important part of
how you apply physics in practice.
The total driving force is then:
$$
F_{D}=F+f_{c} \exp \left(-\left(t / t_{c}\right)^{2}\right)-f_{v} v
$$
where reasonable values for the parameters are \(f_{c}=488 \mathrm{~N}\) and
\(t_{c}=0.67 \mathrm{~s}\). (These values are chosen so that the total race-time
becomes reasonable).
In addition, we need to modify the air resistance force because the runner is
crouched in the initial phase, so that the cross-sectional area is reduced. We
therefore need to replace the cross-sectional area \(A\) in the expression for
\(D\) with a timedependent expression, \(A(t)\), with the properties that: (1)
when time is zero, the area should be reduced to \(75 \%\) of the area during
upright running (again, we guess reasonable values); and (2) after a time much
larger than \(t_{c}\), the runner is upright, and the cross-sectional area
should be \(A\). Again, we introduce a modification to the area that depends on
the exponential factor used above:
$$
A(t)=A-0.25 A \exp \left(-\left(t / t_{c}\right)^{2}\right)=A\left(1-0.25 \exp
\left(-\left(t / t_{c}\right)^{2}\right)\right)
$$
The air resistance force therefore becomes:
\(D=\frac{1}{2} A(t) \rho C_{D}(v-w)^{2}=\) total force on the runner is:
where \(F=400 \mathrm{~N}\) is a constant driving force, and the other terms
have been addressed above.
(i) Modify your numerical method to include these new forces. Find and plot
\(x(t)\), \(v(t)\), and \(a(t)\) for the motion.
(j) How fast does he run \(100 \mathrm{~m}\) ?
(k) Compare the magnitudes of the various forces acting on the runner by
plotting \(F\) (which is constant), \(F_{C}, F_{V}\) and \(D\) as a function of time
for a \(100 \mathrm{~m}\) race. Discuss how important the various effects are.
(1) Use the model to test how the resulting time on \(100 \mathrm{~m}\) would
change if the runner had a hind wind with a wind speed of \(w=1 \mathrm{~m} /
\mathrm{s}\). What if he was running into a wind with a wind speed of \(w=1
\mathrm{~m} / \mathrm{s}\) ?
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