Can someone please show me how to do this?
A hockey player is standing on his skates on a frozen pond when an opposing player moving with a uniform speed of 12m/s skates by with the puck. After 3s, the first player makes up his mind to chase his oppponent. If he accelerates uniformly at 4m/s^2, (a)how long does it take him to catch his opponent and (b)how far has he traveled in this time? (Assume the player with the puck remains in motion at constant speed.)
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To solve this problem, we can use the equations of motion for uniformly accelerated motion. These equations relate the initial velocity (u), final velocity (v), acceleration (a), time (t), and displacement (s).
First, let's find the initial velocity (u) of the first player. We are told that the opposing player is moving with a uniform speed of 12 m/s. Since the first player is standing still initially, his initial velocity is 0 m/s.
Next, we can use the equation s = ut + (1/2)at^2 to find the distance traveled (s) by the first player when he catches his opponent. In this equation, s represents the displacement, u is the initial velocity, t is the time, and a is the acceleration.
For part (a), we need to find the time (t) it takes for the first player to catch his opponent. We can rearrange the equation to solve for t:
s = ut + (1/2)at^2
Since we want to find the time when the first player catches his opponent, we can set the displacement (s) equal to the distance traveled by the opposing player during that time. In this case, since the opposing player is moving with a constant speed of 12 m/s, we can find the distance traveled by multiplying the speed by the time:
s = 12t
Substituting this value into the equation:
12t = 0t + (1/2)(4t^2)
12t = 2t^2
Divide both sides by 2t:
6 = t
Therefore, it takes 6 seconds for the first player to catch his opponent.
For part (b), we need to find the distance traveled (s) by the first player in this time. We can substitute the value of t into the equation s = ut + (1/2)at^2:
s = 0(6) + (1/2)(4)(6^2)
s = 0 + 72
s = 72 meters
Therefore, the first player has traveled a distance of 72 meters when he catches his opponent.