Problem

Using Models Use the model given to answer the qu…

07:29

Problem 10 Medium Difficulty

Using Models Use the model given to answer the questions about the object or process being modeled.

The power $P$ measured in horsepower (hp) needed to drive a
certain ship at a speed of $s$ knots is modeled by
$$P=0.06 s^{3}$$
(a) Find the power needed to drive the ship at 12 knots.
(b) At what speed will a 7.5 -hp engine drive the ship?

Answer

a)
$\begin{aligned} P &=0.06 s^{3} \\ P &=0.06(12)^{3} \\ P &=103.68 \mathrm{hp} \end{aligned}$
b)
$\begin{aligned} P &=0.06 s^{3} \\ 7.5 &=0.06 s^{3} \\ s^{3} &=\frac{7.5}{0.6} \\ s^{3} &=125 \\ s &=5 \mathrm{knots} \end{aligned}$

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Modeling the Real World with Algebra

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Video Transcript

this problem gives us model power equals 0.6 times feed. Cute. So we're going to be using that for both parts of this problem. Part a tells us that if our speed is to be 12 knots, what power are we in need of in order to maintain this speed? So essentially we have r s, and we know that what we're solving for is rp variable. So we're going to plug that into the model that's given to us. So, Pete, to the variable. And then we have P equals 0.6 times s, which is 12 camp. Then we're just going to solve this equation to find what P is. And that will be her answer for part A. So the first step is going to be our exponents. So we get power equals you're a 0.6 times 12 cubed is 1,728 and then we will play this out. We yet a power of 103 0.68 horsepower. It is necessary to move the ship at speed of 12 knots. And there's our answer for apart, eh? So moving on to part B. We're going to dio a very similar approach. But this time we're plugging in our 7.5 horse powers for Pete. And we're leaving s as are variable that we're solving for. So this time the equation becomes 7.5 somewhere substituting for our pee right there equals 0.6 as cubed. So from right now, we're going to isolate our answer variable. So our first step is going to be to divide both sides by I 0.6 So we're left with s cubed equals 21 125. And then finally, to get rid of this explosion on the S, we're going to take the cube root of both sides of the equation and we get our speed is I have not. So there's our solution to party

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James Stewart, Lothar Redlin, Saleem Watson

Algebra and Trigonometry

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