Two soccer players start from rest, 30 m apart. They run directly toward each other, both players accelerating. The first player’s acceleration has a magnitude of 0.49 m/s2. The second player’s acceleration has a magnitude of 0.43 m/s2. (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?
a. d1 + d2 = 30 m.
0.5a1*t^2 + 0.5a2*t^2 = 30.
0.5*0.49*t^2 + 0.5*0.43*t^2 = 30.
t = ?.
b. d1 = 0.5a1*t^2.
To find the time it takes for the players to collide, we can use the kinematic equation:
\[s = ut + \frac{1}{2}at^2\]
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
For the first player, their initial velocity (u) is 0 m/s since they start from rest. The acceleration (a) is given as 0.49 m/s^2. The displacement (s) is 30 m since they start 30 m apart.
For the second player, the initial velocity (u) is also 0 m/s, and the acceleration (a) is given as 0.43 m/s^2. The displacement (s) is -30 m since they start 30 m apart but are running towards each other.
(a) To find the time it takes for the players to collide, we can set up the following equation for the first player:
\[30 = 0 \cdot t + \frac{1}{2} \cdot 0.49 \cdot t^2\]
Simplifying this equation, we get:
\[0.245t^2 = 30\]
Now, we can solve for t:
\[t^2 = \frac{30}{0.245} = 122.45\]
Taking the square root of both sides, we find:
\[t = \sqrt{122.45} \approx 11.06\]
Therefore, it takes approximately 11.06 seconds for the players to collide.
(b) To find how far the first player has run at the instant they collide, we can use the equation:
\[s = ut + \frac{1}{2}at^2\]
For the first player, the time (t) is 11.06 seconds, the initial velocity (u) is 0 m/s, and the acceleration (a) is 0.49 m/s^2. Using these values, we can calculate the displacement:
\[s = 0 \cdot 11.06 + \frac{1}{2} \cdot 0.49 \cdot (11.06)^2\]
Simplifying this, we get:
\[s = 0 + \frac{1}{2} \cdot 0.49 \cdot 122.35\]
\[s = 0 + 30.05\]
Therefore, at the instant they collide, the first player has run approximately 30.05 meters.