Two soccer players start from rest, 25 m apart. They run directly toward each other, both players accelerating. The first player’s acceleration has a magnitude of 0.54 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?

Question

Two soccer players start from rest, 25 m apart. They run directly toward each other, both players accelerating. The first player’s acceleration has a magnitude of 0.54 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?

Answer 1

Use: s(t) = s(0) + v(0) t + (1/2) a t^2

Starting from rest, initial positions 25m apart, with the given accelerations:
s1(t) = (0.54/2) t^2
s2(t) = 25 - (0.40 /2) t^2
Let C be the time of collision:
s1(C) = s2(C)

(A) Solve for C.
(B) Find s1(C).

Answer 2

To find the answers to these questions, we can use the equations of motion from kinematics. The key equation we'll need is:

\(d = ut + \frac{1}{2}at^2\)

Where:
- \(d\) is the displacement
- \(u\) is the initial velocity
- \(a\) is the acceleration
- \(t\) is the time

Let's solve these questions step by step:

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

To find the time before the players collide, we can determine when their displacements become equal. Since they are moving towards each other, their displacements will eventually sum up to the initial separation of 25 meters.

For player 1:
\(d_1 = ut_1 + \frac{1}{2}a_1t_1^2\)

For player 2:
\(d_2 = ut_2 + \frac{1}{2}a_2t_2^2\)

Since they start from rest, their initial velocities (\(u_1\) and \(u_2\)) are zero. The total displacement (\(d_1 + d_2\)) will be 25 meters.

Therefore, we have:
\(\frac{1}{2}a_1t_1^2 + \frac{1}{2}a_2t_2^2 = 25\)

Given \(a_1 = 0.54 m/s^2\) and \(a_2 = -0.40 m/s^2\) (negative because their accelerations are in opposite directions), we use the equations above and solve for \(t_1\) and \(t_2\).

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

To find the distance the first player has run, we can use the equation:

\(d_1 = ut_1 + \frac{1}{2}a_1t_1^2\)

Since the initial velocity (\(u_1\)) is zero and we have already found the time \(t_1\), we can plug in the values and calculate \(d_1\).

So, let's calculate the time (\(t_1\)) and distance (\(d_1\)).