How Fast Did Klaus Spinka Accelerate While Grass Skiing?

[bballgirlweez]

The record speed for grass kiing was set in 1985 by Klaus Spinka, of Austria. Suppose it took Spink 6.6s to reach hs top spped after he started from rest down a slope with a 34.0 degree incline. If teh coefficint of kinetic frction betweent eh skis and the grass was .198, what was the magnitude of Spinka's net acceleration? waht was his speed after 6.60s?

 

[Kelvie]

Please learn to check your spelling, and I think this belongs in the homework help section.

Either way, you should attempt a problem before just posting it (poorly) on a forum, and expecting us to do it for you.

 

[rcgldr]

The record speed for grass skiing was set in 1985 by Klaus Spinka, of Austria. Suppose it took Spink 6.6s to reach his top speed after he started from rest down a slope with a 34.0 degree incline. If teh coefficint of kinetic frction between the skis and the grass was .198, what was the magnitude of Spinka's net acceleration? what was his speed after 6.60s?

This is a terminal velocity problem, and in an ideal situation, the velocity approaches, but never reaches [tex]v_{max}[/tex] The problem needs to be restated so that Spink took 6.6s to reach a certain percentage of his top speed (like 99.5%, 99%, 98%), and these small changes make a big difference.

Here's a link to terminal velocity formula and it's derviation:

http://www.karlscalculus.org/l12.1.html

I changed the signs for Vmax and acceleration so the formula becomes:

[tex]V(t) = V_{max} (1 - e^{-2at/V_{max}})/(1+e^{-2at/V_{max}})[/tex]

where a is the rate of non-aerodynamic acceleration (negative), t is time, V is velocity (0 or negative) and Vmax is max velocity.

In this case, a is the compenent of gravity - component of friction:

[tex]a = (sin(34)\ g) - (cos(34)\ g\ .198) = -9.8m/s^2\ (.559 - .164) = -3.87m/s^2 [/tex]

However, stuck, at [tex]t = 0[/tex], you end up with

[tex] V(0) = V_{max}*(0)/(2) = 0 [/tex]

and at [tex]t = \infty [/tex] you end up with

[tex] V(\infty) = V_{max}*(1)/(1) = V_{max} [/tex]


So as I orignally stated, you can't reach maximum velocity in any finite amount of time in this idealized case. However, if you knew that a certain percentage of Vmax was reached in 6.6 seconds, you could then solve for Vmax. As this percentage approaches 100%, Vmax approaches 0, aerodynamic drag approaches infinity, and vice versa.