### You should know¶

$$ \large L = L_0 \sqrt{1-(v/c)^2} $$

### Solution¶

A rod measures 1.00 m in its rest system. To make it 0.80 m, an observer must move parallel to the rod at a speed of

$$ \large 0.8 = 1.0 \sqrt{1 - (v/c)^2} $$$$ \large 1 - \left( \frac{v}{c} \right)^2 = \left( \frac{4}{5} \right)^2 $$$$ \large \therefore v = 0.60c $$### Answer¶

(B) 0.60$c$