A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 2.0 m/s, skates by with the puck. After 1.00 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 0.20 m/s2, determine each of the following.
(a) How long does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.)
(b) How far has he traveled in that time?
Please explain in great detail step by step if possible I am so lost
a. d1 = V*t=2m/s. * 1s.=2 m. Head start.
d2 = 2 + d1.
0.5a*t^2 = 2 + V1*t.
0.5*0.2*t^2 = 2 + 2*t.
0.1t^2 -2t - 2 = 0.
Use Quadratic Formula.
t = 21 s.
b. d = 2 + 2*21 = 44 m.
No problem! I'll be happy to break down the problem and explain it step by step.
To solve this problem, we need to determine the time it takes for the first player to catch his opponent (part a) and the distance he travels in that time (part b).
Let's start with the information given in the problem:
Player 1's initial speed (standing still) = 0 m/s
Player 2's speed (opponent) = 2.0 m/s
Player 1's acceleration = 0.20 m/s^2
Time interval (when Player 1 starts chasing) = 1.00 s
Now, let's proceed with the solution.
Step 1: Calculate Player 1's speed at the time interval when he starts chasing.
Using the formula: v = u + at
where
v = final velocity,
u = initial velocity (0 m/s),
a = acceleration (0.20 m/s^2),
t = time (1.00 s).
Plugging in the values, we get:
v = 0 + (0.20 * 1.00) = 0.20 m/s
So, Player 1's speed at the start is 0.20 m/s.
Step 2: Calculate the relative speed between Player 1 and Player 2.
Since Player 2's speed remains constant at 2.0 m/s, the relative speed between the two players is the difference between their speeds:
Relative speed = Player 2's speed – Player 1's speed
= 2.0 m/s – 0.20 m/s
= 1.80 m/s
So the relative speed between the players is 1.80 m/s.
Step 3: Determine the time it takes for Player 1 to catch his opponent.
To catch up with the opponent, Player 1 needs to cover the distance between them. Since the players are moving at a relative speed of 1.80 m/s, the time it takes to catch up is the distance divided by the relative speed.
Using the formula: time = distance / speed,
where
time = time taken to catch up,
distance = distance between the players,
speed = relative speed (1.80 m/s).
We don't have the value for distance, so we need to calculate it.
Step 4: Calculate the distance between the players.
To find the distance, we'll use the equation of motion:
s = ut + (1/2)at^2
where
s = distance,
u = initial velocity (0.20 m/s),
t = time (1 s),
a = acceleration (0.20 m/s^2).
Plugging in the values, we get:
s = (0.20 * 1) + (1/2)*(0.20)*(1^2)
s = 0.20 + 0.10 = 0.30 m
Therefore, the distance between the players is 0.30 meters.
Step 5: Calculate the time it takes to catch up.
Now that we have the distance (0.30 m), we can calculate the time using the formula:
time = distance / speed
time = 0.30 m / 1.80 m/s
time ≈ 0.17 s
So, it takes approximately 0.17 seconds for Player 1 to catch his opponent.
Step 6: Calculate the distance traveled by Player 1 in that time.
Since we know the initial velocity (0.20 m/s) and the time it takes to catch up (0.17 s), we can use the equation:
s = ut + (1/2)at^2
where
s = distance,
u = initial velocity (0.20 m/s),
t = time (0.17 s),
a = acceleration (0.20 m/s^2).
Plugging in the values, we get:
s = (0.20 * 0.17) + (1/2)*(0.20)*(0.17^2)
s = 0.034 + 0.00578 ≈ 0.04078 m
Therefore, Player 1 travels approximately 0.04078 meters in that time.
To summarize:
(a) It takes approximately 0.17 seconds for Player 1 to catch his opponent.
(b) Player 1 travels approximately 0.04078 meters in that time.
I hope this detailed explanation helps you understand the problem and the step-by-step solution. Let me know if you have any further questions!