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# (II) When at rest, a spaceship has the form of an isosceles triangle whose two equal sides have length 2$\ell$ and whose base has length $\ell$ . If this ship flies past an observer with a relative velocity of $v=0.95 c$ directed along its base, what are the lengths of the ship's three sides according to theobserver?

## $$\begin{array}{l}{0.31 \ell} \\ {1.94 \ell}\end{array}$$

### Discussion

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##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

##### Jared E.

University of Winnipeg

### Video Transcript

lent off base lend off base is equal to l multiply by square root off one minus B squared, divided by six were And we have we since ah, we is equal to 0.9 five times the speed of light. And therefore, um, best land it's equal to and, uh, into one minus zero point 95 old square. And so lend off. Beth, it's April too. 0.31 times. Now let's find out the horizontal land. So what is onto land is equal to see the point five Zito, do you 0.5 zero times? Oh, l into one minus, uh, 0.95 full square and therefore simplifying this horizontal and physical too. 0.156 times Hello and lent off leg is equal to, uh, square root off horizontal length. Uh, horizontal length square. Ah, bless were to go length, square and, like in the values well, horizontal length is 0.156 and old square bliss vertical limp is one point 93 six times l all square and therefore, uh, lent off leg is equal to 1.94 times and

##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

##### Jared E.

University of Winnipeg