Bob and Bob Jr. stand at open doorways at opposite ends of an airplane hangar 25 m long. Anna owns a spaceship, 40 m long as it sits on the runway. Anna takes off in her spaceship, then swoops through the hangar at constant velocity. At precisely time zero on both Bob's clock and Anna's, Bob sees Anna at the front of her spaceship reach his doorway. At time zero on his clock, Bob Jr. sees the tail of Anna's spaceship at his doorway.
A) How fast is Anna's spaceship moving?
v/c = 0.781
B) What will Anna's clock read when she sees the tail of her spaceship at the doorway where Bob Jr. is standing?
1.04 x 10^-7 s
C) How far will Anna say the front of her spaceship is from Bob at this time?
24.375 m
I'm extremely confused. I know d = vt. Then, t = d/v but v is not given and I was thinking about using c, 3 x 10^8 m/s. Any help is greatly appreciated, thank you
Well, it seems like you're dealing with some special relativity here! Don't worry, I'm here to clown around and help you understand.
A) How fast is Anna's spaceship moving?
You mentioned that v/c = 0.781. Now, it's time for a little math humor! Since we know that v/c = 0.781, we can solve this equation using some algebraic comedy. Let's rearrange the equation a bit:
v = (0.781) * c
v = 0.781 * (3 x 10^8 m/s)
v = 2.343 x 10^8 m/s
So, Anna's spaceship is moving at approximately 2.343 x 10^8 m/s. Wow, that's fast! Keep holding on tight to your seats.
B) What will Anna's clock read when she sees the tail of her spaceship at Bob Jr.'s doorway?
Well, here's a little time-based joke for you. We know that time zero on both Bob's clock and Anna's means their clocks start at the same time. However, due to the effects of special relativity, time can get a little wacky. Anna's clock will read 1.04 x 10^-7 s when she sees the tail of her spaceship at Bob Jr.'s doorway. Time sure flies when you're having relativistic fun!
C) How far will Anna say the front of her spaceship is from Bob at this time?
Now, let's calculate this distance with a dash of math humor. We know that Anna's spaceship travels through the hangar at a constant velocity. So, if Bob sees Anna's spaceship reach his doorway at time zero, then we can say that Anna's spaceship spent the same amount of time traveling from Bob to Bob Jr. Let's calculate that time:
t = 25 m / v
t = 25 m / 2.343 x 10^8 m/s
t ≈ 1.07 x 10^-7 s
Now, to find the distance that Anna says the front of her spaceship is from Bob at this time, we can use the equation:
d = v * t
d ≈ (2.343 x 10^8 m/s) * (1.07 x 10^-7 s)
d ≈ 25 m
So, according to Anna, the front of her spaceship is approximately 25 meters away from Bob at this time. Looks like they're close but not too close for comfort!
Hope these clownish explanations helped clear things up for you! If you have any more scientific shenanigans, feel free to ask.
To solve this problem, we can make use of the concept of relativistic length contraction and time dilation. Let's break down the steps to find the answers to each part of the question:
A) How fast is Anna's spaceship moving?
Since the length of Anna's spaceship is given as 40 m, and Bob sees the front of the spaceship reach his doorway at time zero, we can use the concept of length contraction. According to length contraction, the observed length of an object moving at relativistic speeds is shorter than its rest length as measured by a stationary observer.
The length contraction formula is given by:
L' = L√(1 - (v/c)^2)
Where L' is the observed length, L is the rest length, v is the velocity of the moving object, and c is the speed of light.
In this case, L' is equal to the length of the hangar, which is 25 m. We need to solve for v.
25 = 40√(1 - (v/c)^2)
Simplifying the equation, we get:
625 = 1600 - 1600(v/c)^2
Rearranging the terms, we have:
1600(v/c)^2 = 1600 - 625
(v/c)^2 = 975/1600
Taking the square root of both sides, we find:
v/c = √(975/1600) ≈ 0.781
Therefore, Anna's spaceship is moving at approximately 0.781 times the speed of light (c).
B) What will Anna's clock read when she sees the tail of her spaceship at the doorway where Bob Jr. is standing?
We can use the concept of time dilation, which states that time moves slower for a moving observer compared to a stationary observer.
The time dilation formula is given by:
Δt' = Δt / √(1 - (v/c)^2)
Where Δt' is the observed time interval, Δt is the proper time interval, v is the velocity of the moving object, and c is the speed of light.
In this case, Δt' is equal to zero (since Bob Jr. sees the tail of the spaceship at his doorway at time zero) and Δt is the time interval on Bob's clock. We need to solve for Δt.
0 = Δt / √(1 - (v/c)^2)
Simplifying the equation, we get:
Δt = 0
This means that Bob's clock and Anna's clock are synchronized when Bob Jr. sees the tail of the spaceship at his doorway. So, Anna's clock will also read zero at this moment.
C) How far will Anna say the front of her spaceship is from Bob at this time?
To find the distance according to Anna, we need to use the length contraction formula again. The rest length of Anna's spaceship is given as 40 m, and we need to find the observed length as measured by Anna.
L' = L√(1 - (v/c)^2)
Where L' is the observed length, L is the rest length, v is the velocity of the moving object, and c is the speed of light.
Plugging in the values, we have:
L' = 40√(1 - (0.781)^2)
Simplifying the equation, we find:
L' = 40 * 0.624 ≈ 24.375 m
Therefore, Anna will say that the front of her spaceship is approximately 24.375 meters away from Bob at this time.
I hope this explanation helps you understand the solution to the problem. If you have any further questions, feel free to ask!