# Downhill Ski Race: Step-by-Step Solution A skier moves **70. m down a 30° slope**, **neglecting friction**. We are to: - () Final speed time taken, starting from rest. (b) Same, but starting with initial speed \( v_ = 2.50\,\mathrm{m/s} \). - (c) Discussion. --- ## **Step 1: Analyze the Problem** - **Displacement along slope:** \( d = 70.\,\mathrm{m} \) - **Angle:** \( \theta = 30^\circ \) - **Initial speed:** \( v_ \) (either or 2.50 m/s) - **Neglect friction** - **Acceleration:** Only gravity component down the slope. ### **Acceleration Down the Slope** The component of gravity along the slope is: \[ a = g \sin\theta = 9.80\,\mathrm{m/s}^2 \times \sin(30^\circ) = 9.80 \times .5 = 4.90\,\mathrm{m/s}^2 \] --- ## **(a) Starting from Rest (\( v_ = \))** ### **Final Speed (\( v_f \))** Use kinematic equation: \[ v_f^2 = v_^2 + 2 a d \] \[ v_f^2 = + 2 \times 4.90 \times 70. = 686 \] \[ v_f = \sqrt{686} \approx 26.2\,\mathrm{m/s} \] ### **Time Taken (\( t \))** \[ d = v_ t + \frac{1}{2} a t^2 \] \[ 70. = + .5 \times 4.90 \times t^2 \] \[ 70. = 2.45 t^2 \] \[ t^2 = \frac{70.}{2.45} \approx 28.57 \] \[ t = \sqrt{28.57} \approx 5.35\,\mathrm{s} \] --- ## **(b) Starting with \( v_ = 2.50\,\mathrm{m/s} \)** ### **Final Speed (\( v_f \))** \[ v_f^2 = v_^2 + 2 a d \] \[ v_f^2 = (2.50)^2 + 2 \times 4.90 \times 70. = 6.25 + 686 = 692.25 \] \[ v_f = \sqrt{692.25} \approx 26.3\,\mathrm{m/s} \] ### **Time Taken (\( t \))** Solve: \[ d = v_ t + \frac{1}{2} a t^2 \] \[ 70. = 2.50 t + 2.45 t^2 \] This is a quadratic equation: \[ 2.45 t^2 + 2.50 t - 70. = \] Use quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 2.45, b = 2.50, c = -70. \): \[ t = \frac{-2.50 \pm \sqrt{(2.50)^2 - 4 \times 2.45 \times (-70.)}}{2 \times 2.45} \] \[ t = \frac{-2.50 \pm \sqrt{6.25 + 686}}{4.90} \] \[ t = \frac{-2.50 \pm \sqrt{692.25}}{4.90} \] \[ t = \frac{-2.50 \pm 26.33}{4.90} \] Take positive root: \[ t = \frac{23.83}{4.90} \approx 4.86\,\mathrm{s} \] --- ## **(c) Discussion** ### **Does the answer surprise you?** - **No:** The final speed and time are not greatly affected by the initial speed because the gain in kinetic energy from gravity (\( mgh \)) is much greater than any reasonable starting kinetic energy (\( \frac{1}{2} mv_^2 \)). - **Numerically:** Even with a running start of 2.5 m/s, final speed increased by only \( .1\,\mathrm{m/s} \) and the time decreased by about .5 s over a 5.35 s run. ### **Why is a running start still advantageous in competitive events?** - **Competitive events are won by small margins:** Even fractions of a second matter. - **Running start shaves off time:** Any reduction, even a few tenths of a second, can decide a race. - **Psychological advantage and momentum:** Can also help with confidence and technique. --- ## **Summary Table** | Case | Final Speed (\(\mathrm{m/s}\)) | Time (\(\mathrm{s}\)) | |------------|-------------------------------|----------------------| | (a) Rest | 26.2 | 5.35 | | (b) 2.5 m/s| 26.3 | 4.86 | --- ## **Final Answers** **(a)** - Final speed: \( \boxed{26.2\,\mathrm{m/s}} \) - Time taken: \( \boxed{5.35\,\mathrm{s}} \) **(b)** - Final speed: \( \boxed{26.3\,\mathrm{m/s}} \) - Time taken: \( \boxed{4.86\,\mathrm{s}} \) **(c)** - The small difference in speed and time is because the gravitational potential energy gained dominates the skier's kinetic energy, making the initial speed relatively unimportant. However, in competitive events, even small time advantages can be decisive.