Respuesta :
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
[tex] t_1 = \frac{d}{v + 50}[/tex]
time taken to reach back
[tex] t_2 = \frac{d}{v - 50}[/tex]
now we have total time
[tex]T = t_1 + t_2[/tex]
[tex]11 = \frac{d}{v - 50} + \frac{d}{v + 50}[/tex]
here d= 3000 miles
[tex]11 = \frac{3000}{v - 50} + \frac{3000}{v + 50}[/tex]
[tex]3.67 * 10^{-3} = \frac{2v}{v^2 - 2500}[/tex]
[tex]v^2 - 2500 = 545.45v[/tex]
solving above quadratic equation we will have
[tex]v = 550 mph[/tex]
so speed of plane will be 550 mph
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.
