# A ball is kicked past a player who has a reaction time before he chases after the ball. The player accelerates to catch up to the ball rolling at constant speed. What is the time and distance the player has to run for after the ball passes the player?

## During a soccer game, a ball is kicked past a player at $24.0 \frac{k m}{h}$. The player, who was initially at rest, has a reaction time of $0.85 s$ before he starts to chase after the ball. The player is able to accelerate at $1.93 \frac{m}{s} ^ 2$, while the ball continues to roll at constant speed. a) How long after the ball passes the player, does the player take to catch up to the ball? b) How far does the player have to run to catch up to the ball?

Apr 5, 2016

I get to 7.7s and 57.2m

#### Explanation:

The ball is moving ar24km//h. We need to get this into m/s:

$24 k m {h}^{-} 1 = \frac{24000}{3600} = 6.67 m {s}^{-} 1$
The player has an initial speed of $u = 0 m {s}^{-} 1$ and acceleration $a = 1.93 m {s}^{-} 2$

The ball will have travelled a distance of $6.67 m {s}^{-} 1 \times 0.85 s = 5.67 m$ before the player moves and we will define the time before the player reaches it as $t$.

For the player, to determine the distance travelled , we can use $s = u t + \frac{1}{2} a {t}^{2} = 0 + \frac{1}{2} \cdot 1.93 \cdot {t}^{2}$

This has to equal the distance the ball has travelled
$5.67 + 6.67 \cdot t$

Combining we arrive at a quadratic equation (groan!)
$\frac{1}{2} \cdot 1.93 {t}^{2} - 6.67 t - 5.67 = 0$

Using the general formula to solve this of

$- b \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

$t = 6.67 \pm \frac{\sqrt{{6.67}^{2} + 4 \cdot \frac{1}{2} \cdot 1.93 \cdot 5.67}}{2 \cdot \frac{1}{2} \cdot 1.93} = 7.7 s \mathmr{and} - 0.8 s$

Clearly t cannot be negative so the answer is 7.7s. Substituting this gives the distance the player has to run as 57m.

However, this would need the player to reach a final velocity of $v = u + a t = 1.93 \times 7.7 s = 15 \frac{m}{s}$ which is faster than Usain Bolt!

Apr 6, 2016

(a) $8.5 s$ rounded to one decimal place.
(b) $56.47 m$ rounded to two decimal places.

#### Explanation:

Let us suppose that the player takes time $t$ after the ball passes to catch up to the ball. Let the time be in seconds.
1. Distance traveled by the ball in this duration

$s = v \times t$, $v$ is the constant speed of the ball.
${s}_{b} = 24 \times t , k m p h$,
Let's change the speed from $k m p h$ to $m {s}^{-} 1$
Now $1 k m p h = {10}^{3} / 3600 , \therefore 24 k m p h = 24 \times {10}^{3} / 3600 = 6. \dot{6} m {s}^{-} 1$
Therefore ${s}_{b} = 6. \dot{6} t$ .....(1)
2. Distance moved by the player after the ball passes him.

Due to his reaction time first $0.85 s$, he remains stationary. Then he starts running. Actual running time of player $\left(t - 0.85\right) s$.
Using $s = u t + \frac{1}{2} a {t}^{2}$,
${s}_{p} = 0 \left(t - 0.85\right) + \frac{1}{2} \times 1.93 {\left(t - 0.85\right)}^{2}$
${s}_{p} = \frac{1}{2} \times 1.93 {\left(t - 0.85\right)}^{2}$ .....(2)
(a) Equating (1) and (2)
$\frac{1}{2} \times 1.93 {\left(t - 0.85\right)}^{2} = 6. \dot{6} t$, simplifying, multiplying both sides with $\frac{2}{1.93}$ and rearranging in terms of a quadratic

$\frac{1}{2} \times 1.93 \left({t}^{2} - 2 \times 0.85 t + 0.7225\right) - 6. \dot{6} t = 0$
$\left({t}^{2} - 1.7 t + 0.7225\right) - 6.9085 t = 0$

${t}^{2} - 8.6085 t + 0.7225 = 0$
Using the general expression for finding the roots of a quadratic we obtain
$t = \frac{8.6085 \pm \sqrt{{\left(8.6085\right)}^{2} - 4 \times 1 \times 0.7225}}{2}$
$t = \frac{8.6085 \pm 8.4390}{2}$
$t = 8.5 \mathmr{and} 0.08$, Ignoring second root as it is less than reaction time of the player.
$t = 8.5$ rounded to first decimal place
(b) Insert in (2)
${s}_{p} = \frac{1}{2} \times 1.93 {\left(8.5 - 0.85\right)}^{2} = 56.47 m$ rounded to two decimal places.