# Length contraction

1. May 29, 2009

### prime-factor

1. The problem statement, all variables and given/known data

1)An ex-student with a new spacecraft 'buzzes' the school oval on sports day at 1.5 * 10^8 m/s. How long does the 100m track appear to him.

2) The sports day began at 9am. The student buzzed the school at an average speed of 0.5c till the presentation ceremony at 3pm. If the clock on the spaceship matched the school time at 9pm, what was the time on the clock when the presentation started.

^^I am not sure I have done these correctly. If someone could check them, that would be great. Thanks very much.

2. Relevant equations

L = Lo sqrt[1-u2/c2]

T = To/sqrt[1-u2/c2]

3. The attempt at a solution

Let L = 100

Beta Factor:

sqrt[1-u2/c2]

sqrt ( 1 - (1.5^2/3^2))

sqrt ( 1 - 0.25)

sqrt ( 0.75)

So, 100/ (sqrt(0.75))

= 115.47 m

2)

Let u = 0.5c

T(earth) = 6hrs

To(ship) = ?

6 = To / sqrt ( 1 - (0.5^2))

6 = To / sqrt(0.75)

To = 6*Ans

To = 4.5 hrs , therefore time on spaceship clock = 9 + 4.5 = 1:30

2. May 29, 2009

### Ilivian

U must consider the observer, or the frame, carefully.

1) "Appear to him", that means the student is the observer. So u must consider he stands still and the school move. So Lo=100m, not L, and u must find L.

L = Lo sqrt[1-u2/c2]= 100.sqrt[1-(1,5/3)^2]=75m.

2)Yep, u r right :D

3. May 29, 2009

### prime-factor

Thank you.

But, I don't get 75m for some reason.

L = sqrt(1 - (1.5^2/3^2)) * 100
L = sqrt(1 - 0.25)*100
L = sqrt(0.75)*100
L = 86.60254038

Where have I gone wrong?

4. May 29, 2009

### Ilivian

No, u r right, I forget the sqrt :D My bad, sorry.