Motorcycle Skid Speed B0E894

1. **State the problem:** A motorcycle skids to a stop on dry concrete, leaving two skid marks measuring $25m$ and $29m$. We need to calculate the motorcycle's speed before braking. 2. **Formula used:** The skid formula is: $$v = \sqrt{2 \cdot g \cdot \mu \cdot d}$$ where: - $v$ is the initial speed before braking, - $g$ is the acceleration due to gravity ($9.8\,m/s^2$), - $\mu$ is the coefficient of friction between tires and road, - $d$ is the skid distance. 3. **Explanation of variables:** - $\mu$ (coefficient of friction) depends on the road surface. For dry concrete, $\mu$ is typically about $0.7$ to $0.8$. We will use $\mu = 0.75$ as an average. - Since there are two skid marks, we take the longer skid distance to represent the total stopping distance, so $d = 29m$. 4. **Calculate speed:** Substitute values: $$v = \sqrt{2 \cdot 9.8 \cdot 0.75 \cdot 29}$$ Calculate inside the square root: $$2 \cdot 9.8 = 19.6$$ $$19.6 \cdot 0.75 = 14.7$$ $$14.7 \cdot 29 = 426.3$$ So: $$v = \sqrt{426.3}$$ 5. **Simplify:** $$v \approx 20.65\, m/s$$ 6. **Convert to km/h:** $$v = 20.65 \times 3.6 = 74.34\, km/h$$ **Final answer:** The motorcycle's speed before braking was approximately $74.3\, km/h$. --- **Why this formula is appropriate:** The formula relates the initial speed to the friction force and skid distance. It assumes the motorcycle decelerates uniformly due to friction until it stops, which fits the scenario of skidding to a stop on a dry concrete road. **Note:** Using the longer skid mark accounts for the full stopping distance, which is necessary for accurate speed calculation.