Graph the function over a one-period interval.
-Ignoring friction, the time, $\mathrm { t }$ (in seconds), required for a block to slide down an inclined plane is given by the formula $t = \sqrt { \frac { 2 b } { g \sin \theta \cos \theta } }$ where $b$ is the length of the base in feet and $g = 32.2$ feet per second is the acceleration of gravity. How long does it take a block to slide down an inclined plane with a base of 12 feet at an angle of $48 ^ { \circ }$ ? Round your answer to three decimal places.
A) $0.866 \mathrm { sec }$
B) $6.947 \mathrm { sec }$
C) $1.224 \mathrm { sec }$
D) $0.353 \mathrm { sec }$

Question

Quizplus · Accepted Answer

To find the time it takes for the block to slide down, substitute $b = 12$ feet, $g = 32.2$ feet/second$^2$, and $\theta = 48^\circ$ into the given formula. First, convert $\theta$ to radians: $48^\circ = \frac{48\pi}{180} = \frac{4\pi}{15}$ radians. Then, calculate $t = \sqrt{\frac{2 \times 12}{32.2 \times \sin(\frac{4\pi}{15}) \times \cos(\frac{4\pi}{15})}}$. Simplifying this expression and rounding to three decimal places gives $t \approx 1.224$ seconds.

Graph the Function Over a One-Period Interval t\mathrm { t }t

Graph the Function Over a One-Period Interval $\mathrm { t }$