# Length contraction

## Homework Statement

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.

## Homework Equations

L = Lo sqrt[1-u2/c2]

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

## 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

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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

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?

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