Q 125. - Precalculus Enhanced with Graphing Utilities

Question

If friction is ignored, the time t (in seconds) required for a block to slide down an inclined plane is given by the function

$t\left(\theta \right)=\sqrt{\frac{2a}{g \sin \theta \cos \theta }}$

where a is the length (in feet) of the base and g  32 feet per second per second is the acceleration due to gravity. How

long does it take a block to slide down an inclined plane with base a = 10 feet when,

(a) $\theta = 30°$

(b) $\theta =45°$

(c) $\theta =60°$

Step-by-Step Solution

Answer

(a) $t\left(\theta \right)=\sqrt{\frac{20}{8\sqrt{3}}}$

  $$\begin{gathered} \left(b\right) t\left(\theta \right)=\sqrt{\frac{20}{8\sqrt{3}}} \\ \left(c\right) t\left(\theta \right)=\sqrt{\frac{5}{4}} \end{gathered}$$

Step 1. Given information .

Consider the given value of a = 10 and g = 32 .

Step 2 . (a) Formula used .

Formula  used to find the value .

$t\left(\theta \right)=\sqrt{\frac{2a}{g \sin \theta \cos \theta }}$

Substitute the value $a=10 ,g=32 ,\theta =30°$ in formula

$$\begin{gathered} t\left(30°\right)=\sqrt{\frac{2\times 10}{32\times {\displaystyle \frac{1}{2}}\times {\displaystyle \frac{\sqrt{3}}{2}}}} \\ t\left(30°\right)=\sqrt{\frac{20}{8\sqrt{3}}} \end{gathered}$$

Step 3. Formula used to find t θ   if θ = 45 °

The formula is $t\left(\theta \right)=\sqrt{\frac{2a}{g \sin \theta \cos \theta }}$

Substitute the values in formula .

$$\begin{gathered} t\left(45°\right)=\sqrt{\frac{2\times 10}{32\times {\displaystyle \frac{\sqrt{3}}{2}}\times {\displaystyle \frac{1}{2}}}} \\ t\left(45°\right)=\sqrt{\frac{20}{8\sqrt{3}}} \end{gathered}$$

Step 4. Formula used to find t θ if ϑ = 60 ° .

The formula is 

$t\left(\theta \right)=\sqrt{\frac{2a}{g \sin \theta \cos \theta }}$

Substitute the value of $a ,g ,\theta$ .

$$\begin{gathered} t\left(60°\right)=\sqrt{\frac{2\times 10}{32\times {\displaystyle \frac{1}{\sqrt{2}}}\times {\displaystyle \frac{1}{\sqrt{2}}}}} \\ t\left(60°\right)=\sqrt{\frac{5}{4}} \end{gathered}$$