1. **Problem statement:** We analyze the kinetic energy formula $$E = \frac{1}{2} m v^2$$ and answer parts a to d. 2. **Formula and rules:** Kinetic energy depends on mass $m$ and speed $v$. The formula shows $E$ is proportional to $m$ and to the square of $v$. 3. **Part a:** For objects traveling at a particular speed $v$, $v$ is constant. So, $$E = \frac{1}{2} m v^2 \propto m$$ This means kinetic energy $E$ is directly proportional to mass $m$ when speed is fixed. 4. **Part b:** For an object with constant mass $m$, $m$ is fixed. So, $$E = \frac{1}{2} m v^2 \propto v^2$$ Kinetic energy $E$ is proportional to the square of the speed $v$. 5. **Part c:** If speed decreases by 10%, new speed is $$v_{new} = 0.9 v$$ New kinetic energy is $$E_{new} = \frac{1}{2} m (0.9 v)^2 = \frac{1}{2} m (0.81 v^2) = 0.81 E$$ So kinetic energy decreases to 81% of original, a 19% decrease. 6. **Part d:** Brakes convert kinetic energy to heat at a constant rate, so energy lost is proportional to stopping distance $d$. Since $E \propto v^2$, the energy to stop the car increases with $v^2$. Therefore, $$d \propto E \propto v^2$$ Stopping distance is proportional to the square of the speed. **Final answers:** - a) $E \propto m$ - b) $E \propto v^2$ - c) $E$ decreases to 81% if $v$ decreases by 10% - d) Stopping distance $d \propto v^2$ because kinetic energy depends on $v^2$ and brakes dissipate energy at a constant rate.