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A ball is rolling over a soccer field with constant velocity $V_{b} > 0$ at an angle of 45° to the goal line. It starts at the corner of the field (distance $d$ beside the middle of the goal line) at the time $t_0 =0$. At the same time a player starts heading towards the goal on a path perpendicular to the goal line. He starts from a position at distance 2d in front of the middle of the goal line) with $ a_p (t) = a_{0}(1-{t\over ß})$

I've solved the problems but the left thing to do is just the initial acceleration. I use integration to find these solutions:

  • Distance covered by the ball $s_b = V_b t$
  • Time when ball is in front of the middle of the goal line $t_A = {\sqrt2 d\over V_b}$ (with distance $s_A={d\over cos 45°}$)
  • Velocity of player $v_p = a_0t(1-{t\over 2ß})$
  • Distance covered by the player $s_p = {a_0t^2\over 2} (1-{t\over 3ß})$
  • Velocity of the player when he reach the point (the point after the ball reach distance $s_A$) $v_{pA} = {a_{0}d\over v_b} (\sqrt2-{d\over v_bß})$ *I'm inserting $t_A$ into $v_p$, am I right?
  • Initial acceleration of the player in order to reach the point (the point after the ball reach distance $s_A$) at the same time as the ball $a_{0}$ = ??

Any ideas would be very appreciated.

edited: sorry I mixed up the calculations.

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closed as unclear what you're asking by Gert, honeste_vivere, ACuriousMind, sammy gerbil, knzhou Jul 28 '16 at 18:06

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    $\begingroup$ $v_p=\int_0^ta_p(t)dt=\int_0^ta_0(1-\frac{t}{\beta})dt=a_0t-a_0\frac{t^2}{2\beta}=a_0t(1-\frac{t}{2\beta})$ $\endgroup$ – Gert Jul 27 '16 at 17:31
  • $\begingroup$ "I'm trying to derivate $v_p$ over $t$". $a_p=\frac{dv_p}{dt}$. So you already have that. The acceleration at $t=0$ is $a_0$. $\endgroup$ – Gert Jul 27 '16 at 17:36
  • $\begingroup$ @Gert but how can I define $a_0$? $\endgroup$ – hello Jul 27 '16 at 18:27
  • $\begingroup$ you can't. It's a given, a known. It's the acceleration at $t=0$, by definition. $\endgroup$ – Gert Jul 27 '16 at 18:36
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    $\begingroup$ Also:$s_p=\int_0^tv_pdt=\frac12 t^2(1-\frac{1}{3\beta}t)$ $\endgroup$ – Gert Jul 27 '16 at 18:41
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I think some errors crept into your calculation. And at any rate - your confusion is around $a_0$ - which is the quantity you need to solve for (the initial acceleration needed to reach the ball at just the right moment).

The player has to cover a distance $d$ in the time that the ball covers a distance $\sqrt{2} d$. The velocity of the ball is $v_b$, so the mean velocity of the player must be $\frac{v_b}{\sqrt{2}}$.

Given the expression for the acceleration, the instantaneous velocity is

$$v(t) = \int_0^t a_0(1-\frac{\tau}{\beta})d\tau\\ = a_0t(1-\frac{t}{2\beta})$$

The average velocity is the total distance divided by the total time:

$$v_{av} = \frac{\int_0^t v(\tau)d\tau}{t}=\frac12 a_0 t\left(1- \frac{t}{3\beta}\right) $$

Which has to be equal to $\frac{v_b}{\sqrt{2}}$. Finally, we know the time it takes the ball to get to this point:

$$t = \frac{\sqrt{2} d}{v_b}$$

Putting it all together, we find an expression for $a_0$:

$$\begin{align} \frac{v_b}{\sqrt{2}} &= \frac12 a_0 \frac{\sqrt{2} d}{v_b}\left(1 - \frac{\sqrt{2} d}{3\beta v_b}\right) \\ a_0 &= \frac{v_b^2}{d\left(1 - \frac{\sqrt2 d}{3\beta v_b}\right)}\\ \end{align}$$

Dimensionally and directionally this makes sense. The faster the ball goes, the greater the required initial acceleration. Also, the more the player tires (small $\beta$), the greater the required initial acceleration. Also, for a sufficiently small value for $ \beta < \frac{\sqrt2 d}{3v_b}$ the initial acceleration would have to be negative (the player starts by running away from the goal, and then quickly turns around and sprints to the ball...). And when $\beta$ becomes very large, the problem reduces to the usual constant acceleration, and you can trivially see that you get the expected answer.

Just to confirm this, I ran a simple numerical simulation in Python. Using a large value of beta (and the value of $a_0$ calculated with the equation above) I get the following plot (plotting the distance to the goal line) for $\beta=100$:

enter image description here

And this when I use $\beta=1$ - as predicted, the player would "run away" then turn around and accelerate towards the ball:

enter image description here

In both cases the intersection happens at 25.0 m (I set $d=25~\rm{m}$, arbitrarily) at $t=3.5~\rm{s}$ - just as you would expect for a ball velocity of $10~\rm{m/s}$

Code used:

# football problem
import numpy as np
import matplotlib.pyplot as plt

d = 25
vb = 10
a0 = 1
beta = 1
dt = 0.001

# equation says
a0 = vb*vb/(d*(1-np.sqrt(2)*d/(3*beta*vb)))

def acc(t):
    return a0*(1-t/beta)

tv = np.arange(0,4,dt)

v = 0
x=2*d
X=[]
V=[]

for ti,t in enumerate(tv):
    vt = v - dt*acc(t)
    x = x + (v+vt)/2*dt
    v = vt
    V.append(v)
    X.append(x)

plt.figure()
plt.plot(tv, X)
X2 = tv * vb / np.sqrt(2)
plt.plot(tv, X2)
plt.show()
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  • $\begingroup$ Sorry, isn't that should be $v(t) = a_0t(1-{t\over 2ß})$ ? $\endgroup$ – hello Jul 27 '16 at 21:10
  • $\begingroup$ @hello - yes, good catch. I will have to edit... $\endgroup$ – Floris Jul 27 '16 at 21:11
  • $\begingroup$ Why do I have to use average velocity (which is my $s_p$ divided by $t$) instead of instantaneous velocity (my $v_p$) ? $\endgroup$ – hello Jul 28 '16 at 12:58
  • $\begingroup$ Because the average over the interval tells us when you will arrive... $\endgroup$ – Floris Jul 28 '16 at 12:59
  • $\begingroup$ Okay, that's one thing that I don't know. Thank you so much for your kindly reply. $\endgroup$ – hello Jul 28 '16 at 13:01

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