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# Skier Accelerations

CLMECH-UJLSYC

The position of a skier on the mountain as a function of time is given by

$$x(t) = 0.50t^2 + 2.0t$$
$$y(t) = -t^2 + 10\cos(0.50t)+500$$
$$z(t) = 0$$

where the x/y plane is parallel to the surface of the hill with the x-direction pointing across the hill and the y-direction pointing up the hill (with the bottom of the hill being a y-position of zero). The z-direction is normal to the hill.

What is the instantaneous acceleration vector of the skier at a time of $t = 10\space s$?

A

$4.6\space m/s^2$ at $77^\circ$ below the positive x-axis.

B

$2.9\space m/s^2$ at $70^\circ$ below the positive x-axis.

C

$23.7\space m/s^2$ at $60^\circ$ below the positive x-axis.

D

$5.5\space m/s^2$ at $45^\circ$ below the positive x-axis.