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A skier goes down a mountain slope. The skier's position on the slope changes with time according to the following formulas:

$x = 0.32 \space t^2 + 1.1 \space t$
$y = - 0.15 \space t^2 - 2.1 \space t + 98$

...where $x$ and $y$ are the skier's horizontal and vertical coordinates with respect to the origin 98 m below the skier's initial position.

All the units are SI.

What is the skier's velocity at time t = 14 s?


$12 \space \frac {m}{s} \space \text {at} \space 58^o \space \text {below the horizontal}$


$16 \space \frac {m}{s} \space \text {at} \space 32^o \space \text {below the horizontal}$


$16 \space \frac {m}{s} \space \text {at} \space 58^o \space \text {below the horizontal}$


$12 \space \frac {m}{s} \space \text {at} \space 32^o \space \text {below the horizontal}$

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