Question

PLEASE HELP                                                                                                                                 A flight across the US takes longer east to west then it does west to east. This is due to the plane having a headwind flying westward (thus slowing it down) and a tailwind flying eastward (thus pushing it along faster). Suppose you are flying from Philadelphia, PA to San Diego, CA. The distance between the cities is 2,400 miles and the plane's speed is 450 mi/hr. What is the speed of the wind?

Complete the chart
Going westward- Distance - Rate- Time
Going Eastward-

Write an equation to represent the total time if the round trip takes 11 hours.

Solve your equation to find the speed of the wind. Round to the nearest mph.
PLEASE HELP!

PLEASE ANSWER CORRECTLY TO GET BRAINLIEST ANSWER

Answer 1

To model and solve our situation we are going to use the equation: $s= \frac{d}{t}$
where
$s$ is speed
$d$ is distance 
$t$ is time 

1. We know that the distance between the cities is 2400 miles, so $d=2400$. We also know that the speed of the plane is 450 mi/h. Since we don't know the speed of the air, $S_{a}=?$. We don't know how much the westward trip takes, so $t_{w}=?$, and we also don't know how much the eastward trip takes, so $t_{e}=?$.

Going westward. Here the plane is flying against the air, so we need to subtract the speed of the air from the speed of the plane:
$450-S_{a}= \frac{2400}{t_{w} }$
Going eastward. Here the plane is flying with the the air, so we need to add the speed of the air to the speed of the plane:
$450+S_{a}= \frac{2400}{t_{e} }$

2. We know for our problem that the round trip takes 11 hours; so the total time of the trip is 11, $t_{t}=11$. Notice that we also know that the total time of the trip equals time of the tip going westward plus time of the trip going eastward, so $t_{t}=t_{w}+t_{e}$. Since we know that the total trip takes 11 hours, we can replace that value in our total time equation and solve for $t_{w}$:
$11=t_{w}+t_{e}$
$t_{w}=11-t_{e}$

Now we can replace $t_{w}$ in our going westward equation to model our round trip with a system of equations:
$450-S_{a}= \frac{2400}{t_{w}}$
$450-S_{a}= \frac{2400}{11-t_{e} }$ equation (1)
$450+S_{a}= \frac{2400}{t_{e}}$ equation (2)

3. To solve our system of equations, we are going to solve for $t_{e}$ in equations (1) (2):

From equation (1)
$450-S_{a}= \frac{2400}{11-t_{e} }$
$11-t_{e}= \frac{2400}{450-S_{a} }$
$-t_{e}= \frac{2400}{450-S_{a} } -11$
$t_{e}=11- \frac{2400}{450-S_{a} }$
$t_{e}= \frac{4950-11S_{a} -2400}{450-S_{a} }$
$t_{e}= \frac{2550-11S_{a} }{450-S_{a} }$ equation (3)

From equation (2):
$450+S_{a}= \frac{2400}{t_{e} }$
$t_{e}= \frac{2400}{450+S_{a} }$ equation (4)

Replacing (4) in (3)
$\frac{2400}{450+S_{a}} = \frac{2550-11S_{a}}{450-S_{a} }$
Now, we can solve for $S_{a}$ to find the speed of the wind:
$2400(450-S_{a})=(450+S_{a})(2550-11S_{a})$
$1080000-2400S_{a}=1147500-4950S_{a}+2550S_{a}-11(S_{a})^{2}$
$11(S_{a})^{2}-67500=0$
$11(S_{a})^{2}=67500$
$(S_{a})^{2}= \frac{67500}{11}$
$S_{a}=+/- \sqrt{ \frac{67500}{11} }$
Since speed cannot be negative, the solution of our equation is:
$S_{a}= \sqrt{ \frac{67500}{11} }$
$S_{a}=78.33$

We can conclude that the speed of the wind is 78 mph.