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 2.00 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 0.10 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?
I am confused on how to get 2 relating equations to find the time.
The go the same distance.
distance1=2(t)
distance2=1/2 *.1(t-2)^2
set them equal, solve fodr time t. This is the time after initial passing.
solve for distance.
I solved for time, did the quadratic & got 43.9 seconds. I used that to get a distance of 96.4 meters. The program said that they are both incorrect (might be due to significant figures) so I'm not sure where I went wrong.
To solve this problem, you can use the equations of motion. There are two relevant equations that you can use:
1. vf = vi + at
This equation relates the final velocity (vf), initial velocity (vi), acceleration (a), and time (t).
2. d = vit + (1/2)at^2
This equation relates the displacement (d), initial velocity (vi), acceleration (a), and time (t).
In this case, let's assume the initial velocity of the first player (chasing player) to be zero (since he is initially stationary). The velocity of the player with the puck is given as 2.0 m/s, and the acceleration of the chasing player is 0.10 m/s^2.
(a) How long does it take him to catch his opponent?
Using equation 1, we can substitute the known values into the equation:
vf = vi + at
0 = (2.0 m/s) + (0.10 m/s^2)(t)
Now, solve for t:
0.10 m/s^2(t) = -2.0 m/s
t = -2.0 m/s / 0.10 m/s^2
t = -20 s/m^2
Since time cannot be negative in this context, we can conclude that the chasing player will never catch his opponent. There is no solution for this part of the question.
(b) How far has he traveled in that time?
Since the chasing player cannot catch his opponent, we do not need to calculate the distance traveled.
To solve this problem, we can use two key equations: the equation of motion for the first player and the equation for the relative velocity between the two players.
Let's take a look at the equation of motion for the first player:
1. Equation of motion: s = ut + (1/2)at^2
where:
s = distance traveled
u = initial velocity
t = time
a = acceleration
In this case, the initial velocity (u) of the first player is 0 m/s since he was initially stationary. The acceleration (a) is given as 0.10 m/s^2.
We need to find the time it takes for the first player to catch his opponent, so we set up the equation for the relative velocity between the two players:
2. Relative velocity equation: v_rel = v_opponent - v_first_player
where:
v_rel = relative velocity between the two players
v_opponent = velocity of the opponent (2.0 m/s)
v_first_player = velocity of the first player (initially 0 m/s, but increasing with time due to acceleration)
At the moment when the first player starts chasing his opponent (after 2.00 seconds), the opponent has already traveled a distance given by:
distance_opponent = v_opponent * (2.00 s)
To find out when the first player catches his opponent, we set up the equation:
distance_opponent = distance_first_player
v_opponent * (2.00 s) = v_first_player * t
Substituting the values we have:
2.0 m/s * (2.00 s) = 0 + (0.10 m/s^2) * t
From here, you can solve for the time (t). Simplifying the equation gives:
4.0 m = 0.10 m/s^2 * t
Now, divide both sides of the equation by 0.10 m/s^2:
t = 4.0 m / 0.10 m/s^2
t = 40 s
So, it will take the first player 40 seconds to catch his opponent.
To find the distance traveled by the first player in that time, we can use the equation of motion again:
s = ut + (1/2)at^2
Substituting the given values:
s = 0 * (40 s) + (1/2) * (0.10 m/s^2) * (40 s)^2
You can solve this equation to find the distance traveled (s).