Two soccer players start from rest, 32 m apart. They run directly toward each other, both players accelerating. The first player’s acceleration has a magnitude of 0.55 m/s2. The second player’s acceleration has a magnitude of 0.40 m/s2. (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?
To solve this problem, we can use the equations of motion to determine the time it takes for the players to collide and the distance the first player has run at that instant.
Step 1: Determine the time it takes for the players to collide.
Let's assume the initial position of the first player as x1 = 0 m and the initial position of the second player as x2 = 32 m. The equation we will use is:
x2 = x1 + v1t + (1/2) a1t^2 + v2t + (1/2) a2t^2
Since both players start from rest, their initial velocities are 0 m/s, and the equation simplifies to:
32 = 0 + 0 + (1/2)(0.55)t^2 + 0 + (1/2)(-0.40)t^2
Simplifying further:
32 = (0.275)t^2 - (0.20)t^2
Combining like terms:
32 = (0.075)t^2
Dividing by 0.075:
t^2 = 32 / 0.075
t^2 = 426.67
Taking the square root of both sides:
t ≈ √426.67
t ≈ 20.64 s
Therefore, the time it takes for the players to collide is approximately 20.64 seconds.
Step 2: Determine the distance the first player has run when the collision occurs.
We can use the equation for displacement:
x = x1 + v1t + (1/2) a1t^2
Substituting the values, we get:
x = 0 + (0)t + (1/2)(0.55)(20.64)^2
Simplifying:
x ≈ 0 + 0 + (1/2)(0.55)(426.67)
x ≈ 11,766.835 m
Therefore, at the instant they collide, the first player has run approximately 11,766.835 meters.
To find the time it takes for the players to collide and the distance the first player runs before the collision, we can use the equations of motion. Let's break down the problem step by step:
(a) How much time passes before the players collide?
To find the time it takes for the players to collide, we need to determine when their positions will be the same. We'll use the equation:
s = ut + (1/2)at²
where:
s is the distance traveled,
u is the initial velocity,
t is the time taken,
a is the acceleration.
For the first player:
The initial distance traveled (s1) is 0 meters because they start from rest.
The initial velocity (u1) is also 0 m/s because they start from rest.
The acceleration (a1) is given as 0.55 m/s².
For the second player:
The initial distance traveled (s2) is 32 meters because they start 32 m apart.
The initial velocity (u2) is 0 m/s because they start from rest.
The acceleration (a2) is given as 0.40 m/s².
Since we want to find the time before they collide, we set s1 = s2 and solve for t:
0 + (1/2)(0.55)t² = 32 + (1/2)(0.40)t²
Simplifying the equation:
(1/2)(0.55)t² - (1/2)(0.40)t² = 32
(0.275)t² - (0.20)t² = 32
0.075t² = 32
Now, divide both sides of the equation by 0.075:
t² = 32 / 0.075
t² ≈ 426.67
t ≈ √426.67
t ≈ 20.65 seconds (rounded to two decimal places)
Therefore, it takes approximately 20.65 seconds for the players to collide.
(b) At the instant they collide, how far has the first player run?
To find how far the first player has run when they collide, we can substitute the calculated time (t ≈ 20.65 s) into the equation for the displacement:
s = ut + (1/2)at²
For the first player:
s1 = u1t + (1/2)a1t²
Substituting the given values:
s1 = 0 + (1/2)(0.55)(20.65)²
Simplifying the equation:
s1 ≈ 0 + 0.5(0.55)(427.0875)
s1 ≈ 0.5(234.898125)
s1 ≈ 117.4490625 meters
Therefore, at the instant they collide, the first player has run approximately 117.45 meters.