## College Physics (4th Edition)

(a) As measured in the moving reference frame, the lengths of the sides are 30 cm and 58 cm (b) $v = 0.866~c$ $v = 2.6\times 10^8~m/s$
(a) Let $L_0 = 60~cm$ We can find the length $L$ according to the moving reference frame: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $L = (60~cm)~\sqrt{1-\frac{(0.25~c)^2}{c^2}}$ $L = (60~cm)~\sqrt{1-\frac{1}{16}}$ $L =58~cm$ Since the 30-cm side is perpendicular to the direction of motion, there is no length contraction of this side. As measured in the moving reference frame, the lengths of the sides are 30 cm and 58 cm (b) Let $L_0 = 60~cm$ and let $L = 30~cm$ We can find the required speed $v$ of a moving reference frame: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{L}{L_0}$ $1-\frac{v^2}{c^2} = (\frac{L}{L_0})^2$ $\frac{v^2}{c^2} = 1-(\frac{L}{L_0})^2$ $v^2 = [1-(\frac{L}{L_0})^2]~(c^2)$ $v = \sqrt{1-(\frac{L}{L_0})^2}~\times c$ $v = \sqrt{1-(\frac{30~cm}{60~cm})^2}~\times c$ $v = 0.866~c$ $v = (0.866)~(3.0\times 10^8~m/s)$ $v = 2.6\times 10^8~m/s$