Imagine that a person is holding a basket of baseballs, and is throwing them at you, a helpless person without a baseball glove, with a speed of 10m/s, every two seconds. Assume you are a distance of 50m away.
A. How long will it take until you get hit with the first ball?
B. How often will you get pelted with baseball?
Now, Imagine that the attacker is moving towards you with a speed of 5m/s, but can only throw the baseballs with a speed of 10m/s (from your reference frame). The first ball is thrown when the person is 50m from you.
A. How long will it take until you get hit with the first ball?
B. How often will you get pelted with baseballs?
Finally, you get your wits about you and start running away from the mad attacker, at a speed of 5m/s. The attacker was tired and stopped to catch his breath, but started throwing the baseballs again when you are a distance of 50m away from him.
A. How long will it take until you get hit with the first ball?
B. How often will you get pelted with baseballs?
To answer these questions, we need to know the distance between you and the person throwing the baseballs, as well as the speeds involved. Let's break down each scenario.
Scenario 1:
A. The person is throwing baseballs at a speed of 10m/s, and you are 50m away from them. Since the distance covered in the time it takes for the first ball to reach you is equal to the initial distance between you and the person throwing the baseballs, we can use the formula distance = speed * time. In this case, distance = 10m/s * time. Rearranging the formula to solve for time, we get time = distance / speed. Plugging in the values, time = 50m / 10m/s = 5 seconds.
B. The person is throwing the baseballs every 2 seconds. So, you will get pelted with a baseball every 2 seconds.
Scenario 2:
A. In this scenario, the person throwing the baseballs is moving towards you at a speed of 5m/s. The first ball is thrown when the person is 50m away from you. The effective velocity of the baseball thrown would be the sum of the person's throwing speed (10m/s) and their movement speed towards you (5m/s). So, the relative velocity of the baseballs is 10m/s + 5m/s = 15m/s. Using the same formula as before, distance = speed * time, we have distance = 15m/s * time. Rearranging the formula, we get time = distance / speed. Plugging in the values, time = 50m / 15m/s = 3.33 seconds (rounded to two decimal places).
B. The person is still throwing the baseballs every 2 seconds. So, you will get pelted with a baseball approximately every 2 seconds.
Scenario 3:
A. You are now running away from the attacker at a speed of 5m/s. The first ball is thrown when you are 50m away from the attacker. Since you are running away, the effective distance between you and the attacker is the initial distance minus the distance covered by you while the ball is in the air. The distance you cover is your running speed multiplied by the time it takes for the ball to reach you. So, the new distance is 50m - (5m/s * time). Setting this equal to zero (as you would get hit when the distance is zero), we get 50m - (5m/s * time) = 0. Solving for time, we get time = 50m / (5m/s) = 10 seconds.
B. The person is still throwing the baseballs every 2 seconds. However, since you are now running away, it will take longer for each subsequent ball to reach you. So, the frequency at which you get pelted with baseballs will be less than every 2 seconds.