Light, dry snow is called powder. Skiing on a powder day is different than skiin

Question

Light, dry snow is called powder. Skiing on a powder day is different than skiing on a day when the snow is wet and heavy. When you slow down on DRY SNOW the maximum (negative) acceleration caused by the snow acting on your skis is about two-fifths as much as that of stopping on WET SNOW.

1) For a given initial velocity, how does the time t.sub.d it takes to stop on dry snow differ from the time t.sub.w it takes to stop on wet snow?

A. t.sub.d = 0.4 t.sub.w
B. t.sub.d = t.sub.w
C. t.sub.d = 2.5 t.sub.w

2) For a given initial velocity, how does the stopping distance x_d on dry snow differ from the stopping distance x_w on wet snow?

A. x.sub.d = 0.4 x.sub.w
B. x.sub.d = x.sub.w
C. x.sub.d = 2.5 x.sub.w

Explanation / Answer

I think you got a negative result b/c you might have forgotten that the acceleration is negative. Part A: V = u + at_dry I picked u = 3m/s 0 = (3m/s) + (2/5)(-9.8m/s^2) t_dry isolate t_dry and you get t_dry = (-3m/s)/(-3.92m/s^2) = .765s Now do the same for t_wet same u so you get: 0 = (3m/s) + (-9.8m/s^2) t_wet isolate t_wet: t_wet = (-3m/s)/(-9.8m/s^2) = .306s to find the ratio between them take t_dry/t_wet t_dry = 2.5t_wet Part B use X_d = X_start + u t_dry + 1/2 a (t_dry)^2 X_d = 0 + (3m/s)(.765s) + 1/2(-3.92m/s^2)(.765s)^2 X_d = 1.147m use X_w = X_start + u t_wet + 1/2 a (t_wet)^2 X_w = 0 + (3m/s)(.306s) + 1/2(-9.8m/s^2)(.306s)^2 X_w = .459m find the ration the same way in part A: X_d/X_w X_d = 2.5X_w

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