How fast would a 50-meter spaceship (as measured in the rest frame)have to travel to appear 35 meter to a stationary observer?
A)0.25 c
B)0.60 c
C)0.71 c
D)1.0 c
E)1.23 c

Question

Quizplus · Accepted Answer

To solve this problem, we can use the equation for length contraction, which is:
L = L0/γ
where L0 is the rest length of the spaceship (50 meters in this case), L is its length as observed by the stationary observer (35 meters in this case), and γ is the Lorentz factor, given by:
γ = 1/√(1 - v²/c²)
where v is the velocity of the spaceship and c is the speed of light.

Plugging in the values given in the problem, we get:
35 = 50/γ
γ = 50/35 = 10/7

Solving the equation for γ, we get:
γ = 1/√(1 - v²/c²)
(10/7)² = 1 - v²/c²
100/49 = 1 - v²/c²
v²/c² = 1 - 100/49
v²/c² = 51/49
v/c = √(51/49)
v/c ≈ 1.01

Therefore, the speed of the spaceship as measured by the stationary observer is v ≈ 1.01c. The closest choice to this answer is C) 0.71c, which is the correct answer.

How Fast Would a 50-Meter Spaceship (As Measured in the Rest