Question

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

Answer 1

Answer:

550 mph

Explanation:

Let the plane's speed be p. With the wind helping, the plane travels at $p+50$ mph relative to the ground. So, it takes $\dfrac{3000}{p+50}$ hours to make the trip to Halifax. Using similar logic, the way back takes $\dfrac{3000}{p-50}$ hours. Combining these with the 2 hours needed to give the talk yields the total time of 13 hours: \[\dfrac{3000}{p+50} + 2 + \dfrac{3000}{p-50}=13.\]Subtracting 2 and multiplying both sides by $(p+50)(p-50)$ gives: \[3000(p-50)+3000(p+50)=11(p^2-2500)\]This simplifies to $11p^2-6000p-27500=0$. This factors as $(p-550)(11p+50) =0 $, so since $p$ must be positive we have the solution $p=\boxed{550} \text{ mph}$.

You might also simply use a little trial and error on the main time equation above. The two fractions will be roughly equal and must contribute 11 total hours to the left hand side. When $p$ is near $500$, the fractions are about $6$ each; so, trying a few values of $p$ quickly gets to the answer with no messy algebra.

Answer 2

When plane is going towards Halifax the speed of wind is in the direction of fly

so overall the net speed of the plane will increase

while when he is on the way back the air is opposite to flight so net speed will decrease

now the total time of the journey is 13 hours

out of this 2 hours he spent in mathematics talk

so total time of the fly is 13 - 2 = 11 hours

now we have formula to find the time to travel to Halinex

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

time taken to reach back

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

now we have total time

$T = t_1 + t_2$

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

here d= 3000 miles

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

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

$v^2 - 2500 = 545.45v$

solving above quadratic equation we will have

$v = 550 mph$

so speed of plane will be 550 mph