A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 12 m/s, skates by with the puck. After 2.0 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.0 m/s2, determine each of the following.

(a) How long does it take him to catch his opponent?(b) How far has he traveled in that time?

To answer this question, we can use the kinematics equations of motion. We will consider the motion of both the chasing player and the opponent.

First, let's find the speed at which the first player will catch up with the opponent. We'll assume that he catches up when their positions are equal.

(a) To find the time it takes for the first player to catch his opponent, we'll set up equations for their displacements:

For the first player:
His initial position is 0 since he is standing on the frozen pond.
His final position is the distance he covers, which we need to find.
The time it takes for him to catch the opponent is represented by t.

For the opponent:
His initial position is the distance he has already traveled, which is 12 m/s * 2 s.
His final position is the distance covered by the first player, which we need to find.
The time it takes for the first player to catch him is also represented by t.

Setting up the equations:
First player's displacement = opponent's displacement
0 + (1/2) * 4 * t^2 = 12 * 2 + 12 * t
2t^2 = 24t + 24
2t^2 - 24t - 24 = 0

Using the quadratic formula, we can find the value of t:
t = (-(-24) ± sqrt((-24)^2 - 4 * 2 * (-24))) / (2 * 2)
t = (24 ± sqrt(576 + 192)) / 4
t = (24 ± sqrt(768)) / 4
t ≈ (24 ± 27.71) / 4

The solutions to this quadratic equation are t ≈ 13.71 s and t ≈ -1.71 s. We discard the negative solution since time cannot be negative.

Therefore, it takes the first player approximately 13.71 seconds to catch his opponent.

(b) To find how far the first player has traveled at this point, we can substitute the value of t back into either of the displacement equations:

First player's displacement = 0 + (1/2) * 4 * (13.71)^2
= 0 + 0.5 * 4 * 188.4341
= 377.8682 meters

Therefore, the first player will have traveled approximately 377.87 meters by the time he catches his opponent.

To solve this problem, we can use the equations of motion. We'll start by finding the initial speed of the first player, and then we'll use that along with his acceleration to find the time and distance.

(a) To find how long it takes the first player to catch his opponent, we need to determine the initial speed of the first player. We can use the equation:

v = u + at

where:
v = final velocity (12 m/s, the speed of the opponent)
u = initial velocity (unknown)
a = acceleration (4.0 m/s^2)
t = time (2.0 s)

Rearranging the equation to solve for u:

u = v - at

Substituting the known values:

u = 12 m/s - (4.0 m/s^2)(2.0 s)
u = 12 m/s - 8.0 m/s
u = 4.0 m/s

So the initial speed of the first player is 4.0 m/s.

Now, to find how long it takes the first player to catch his opponent, we can use the equation:

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

where:
s = distance travelled (unknown)
u = initial velocity (4.0 m/s)
a = acceleration (4.0 m/s^2)
t = time (unknown)

Since we want to find the time, we rearrange the equation:

t = (-u ± sqrt(u^2 - 2as)) / a

Substituting the known values:

t = (-4.0 m/s ± sqrt((4.0 m/s)^2 - 2(4.0 m/s^2)(0 m))) / (4.0 m/s^2)

Since the term inside the square root is zero, we have:

t = (-4.0 m/s ± 0) / (4.0 m/s^2)
t = -4.0 m/s / 4.0 m/s^2
t = -1.0 s

Since time cannot be negative in this context, we discard the negative solution.

Therefore, it takes the first player 1.0 second to catch his opponent.

(b) To find how far the first player has traveled in that time, we can use the equation:

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

where:
s = distance travelled (unknown)
u = initial velocity (4.0 m/s)
a = acceleration (4.0 m/s^2)
t = time (1.0 s)

Substituting the known values:

s = (4.0 m/s)(1.0 s) + (1/2)(4.0 m/s^2)(1.0 s)^2
s = 4.0 m + (1/2)(4.0 m/s^2)(1.0 s^2)
s = 4.0 m + 2.0 m
s = 6.0 m

Therefore, the first player has traveled 6.0 meters in that time.

a. d1 = Vo*t + at^2,

d2 = 12(t+2),

d1 = d2,

0 + 0.5*4t^2 = 12(t+2),
2t^2 = 12t + 24,
2t^2 - 12t - 24 = 0,
Divide both sides by 2:
t^2 - 6t - 12 = 0.
Use Quadratic Formula to find t:
t = 7.58s.

b. d = 0 + 0.5*4*(7.58)^2 = 114.9s.