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kvasek
1 month ago
6

Richard needs to fly from san diego to halifax, nova scotia and back in order to give an important talk about mathematics. on th

e way to halifax, he will get a speed boost from the wind which blows at 50 miles per hour (mph). on the way back, the plane must, unfortunately, fight this wind speed. if the talk lasts 2 hours, and if the distance between the two cities is 3000 miles, how fast must the plane fly in mph if the entire trip is to take 13 hours?
Physics
2 answers:
kicyunya [3.2K]1 month ago
4 0

Response:

550 mph

Clarification:

Let the plane's speed be represented by p. Under wind assistance, the plane travels at $p+50$ mph in relation to the ground. Therefore, traveling to Halifax takes $\dfrac{3000}{p+50}$ hours. Conversely, the return trip requires $\dfrac{3000}{p-50}$ hours. Adding these to the 2 hours for the talk gives a total of 13 hours: \[\dfrac{3000}{p+50} + 2 + \dfrac{3000}{p-50}=13.\] After subtracting 2 and multiplying both sides by $(p+50)(p-50)$, we arrive at: \[3000(p-50)+3000(p+50)=11(p^2-2500)\]Which simplifies to $11p^2-6000p-27500=0$. Factoring this gives $(p-550)(11p+50) =0 $, leading us to the solution $p=\boxed{550} \text{ mph}$.

Alternatively, trial and error with the main time equation can be effective. The two fractions will be roughly equal, contributing a total of 11 hours. Testing values close to $500$ for p quickly leads to the correct answer without complicated calculations.

Ostrovityanka [3.2K]1 month ago
3 0

As the plane heads toward Halifax, the wind speed supports the flight path

resulting in an overall improved speed

Conversely, during the return trip, the wind will resist the plane's motion, decreasing the net speed

The total journey lasts 13 hours

of which 2 hours was dedicated to the mathematics discussion

Consequently, the total flight time is 13 - 2 = 11 hours

Now we apply the formula to calculate the time for traveling to Halifax

t_1 = \frac{d}{v + 50}

Time needed to return

t_2 = \frac{d}{v - 50}

Let’s look at the total time

T = t_1 + t_2

11 = \frac{d}{v - 50} + \frac{d}{v + 50}

Here d = 3000 miles

11 = \frac{3000}{v - 50} + \frac{3000}{v + 50}

3.67 * 10^{-3} = \frac{2v}{v^2 - 2500}

v^2 - 2500 = 545.45v

By solving the derived quadratic equation

v = 550 mph

the plane's speed calculates to 550 mph

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