Two soccer players start from rest, 41 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 0.43 m/s2. (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?

Question

Two soccer players start from rest, 41 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 0.43 m/s2. (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?

Answer 1

a. a1 = 0.55m/s^2, a2 = 0.43 m/s^2.

d1 + d2 = 41m.

0.5a1*t^2 + 0.5a*t^2 = 41 m
0.275t^2 + 0.215t^2 = 41
0.49t^2 = 41
t^2 = 83.67
t = 9.15 s.

b. d1 = 0.275*(9.15(^2 = 23.0 m.

Answer 2

To solve this problem, we can use the equations of motion.

(a) To find the time it takes for the players to collide, we can use the equation:

s = ut + (1/2)at^2

where s is the distance traveled, u is the initial velocity (which is 0 since the players start from rest), a is the acceleration, and t is the time.

Let's denote the first player as P1 and the second player as P2.

For P1:
s1 = ut1 + (1/2)a1t1^2

For P2:
s2 = ut2 + (1/2)a2t2^2

Since the players are running towards each other, their displacements will add up to the total distance (41 m):
s1 + s2 = 41

Now, let's substitute the values we know into the equations.

For P1:
s1 = 0 + (1/2)(0.55)t1^2

For P2:
s2 = 0 + (1/2)(0.43)t2^2

We already know that s1 + s2 = 41, so we can substitute these values:

(1/2)(0.55)t1^2 + (1/2)(0.43)t2^2 = 41

Now, we need to solve this equation to find the values of t1 and t2.

(b) To find how far the first player has run at the instant they collide, we can use the equation:

s = ut + (1/2)at^2

Since we are interested in the position when the players collide, we can consider the displacement of the first player:

s1 = ut1 + (1/2)a1t1^2

We can substitute the known values:

s1 = 0 + (1/2)(0.55)t1^2

Now, we can calculate the value of s1 when we find the value of t1.

To find the solution to these equations, you can either solve them algebraically or use numerical methods such as graphing or numerical approximation techniques.

Answer 3

To solve this problem, we can use the equation of motion for each player. The equation is given by:

s = ut + (1/2)at^2

Where:
s = distance traveled
u = initial velocity (in this case, the initial velocity is 0 since the players start from rest)
a = acceleration
t = time

Let's solve the problem step-by-step:

(a) How much time passes before the players collide?

We know that the players are initially 41 m apart, and they are moving towards each other. Their relative speed will be the sum of their individual speeds.

Relative speed = (acceleration of player 1) + (acceleration of player 2)
Relative speed = 0.55 m/s^2 + 0.43 m/s^2
Relative speed = 0.98 m/s^2

Now, let's find the time it takes for the players to collide. We can use the equation of motion for both players:

For player 1:
41 = 0*t + (1/2)*(0.55)*t^2

For player 2:
41 = 0*t + (1/2)*(0.43)*t^2

Since both equations are the same, we can solve for t:

(1/2)*(0.55)*t^2 = 41

0.275t^2 = 41

t^2 = 41 / 0.275

t^2 = 149.09

Taking the square root of both sides:

t = √149.09

t ≈ 12.20 seconds

Therefore, it takes approximately 12.20 seconds for the players to collide.

(b) At the instant they collide, how far has the first player run?

To find the distance traveled by the first player at the instant of collision, we can substitute the value of t we just found into the equation of motion for player 1:

s = 0*t + (1/2)*(0.55)*t^2

s = 0 + (1/2)*(0.55)*(12.20^2)

s ≈ 42.60 meters

Therefore, at the instant they collide, the first player has run approximately 42.60 meters.