Verified answer

Ooh, this one's kinda tricky.

Okay, we have two players, A and B. B speeds past A at 12 m/s. After 3.0 seconds, player B is 36 meters away from A.

So we need to find the time required for B to travel the same distance as A, keeping in mind that B already has a 36 meter lead on him. That means we need to define the movements of players A and B in terms of displacement.

For player A:

dx = vo*t + 1/2at²

Since player A's original velocity was zero, we can simplify to:

dx = 1/2at²

For player B:

dx = 36 m + vt

Now just set the two equations equal to each other (since, by definition, when player A catches player B, their displacements from the starting point will be the same) and solve for t.

1/2at² = 36 m/s + vt

1/2(4.0 m/s²)t² = 36 m/s + (12 m/s)t

(2.0 m/s²)t² = 36 m/s + (12 m/s)t

You can rearrange this equation into a proper second-order polynomial:

(2.0 m/s²)t² - (12 m/s)t - 36 m = 0

And use the quadratic formula to solve for t. After doing so, you get two roots, one of which is negative. Since a negative time doesn't make any sense, you can discard it and keep the positive root: 8.2 seconds.

d = 12(3) = 36 m

Distance balance between the two players,

36 + 12t, the first = 0 + (1/2)at^2, the second = (1/2)(4)t^2

Collect all the terms in one side,

t^2-6t-18 = 0

Solve for t with quadratic formula,

t = 3+√27 = 8.2 sec

First player e= 12*t+36

2nd player e= 1/2*4 *t^2

2t^2=12t+36

t^2-6t-18=0

t=((6+sqrt(36+72))/2= 8.2 s