Question

A plane flies from Penthaven to Jackson and then back to Penthaven. When there is no wind, the round trip takes 6 hours and 24 minutes, but when there is a wind blowing from Penthaven to Jackson at 50 miles per hour, the trip takes 6 hours and 40 minutes. How many miles is the distance from Penthaven to Jackson?

(Assume that the plane flies at a constant speed, and that the turnaround time is negligible.)

Answer 1

Answer:800 miles

Step-by-step explanation:

Given

Round trip for Penthaven to Jackson takes 6 hr and 24 minutes in absence of wind

$t_1=6+\frac{24}{60}=6.4\ hr$

When Wind blows from Penthaven to Jackson it takes 6 hr and 40 min i.e.

$t_2=6+\frac{40}{60}=\frac{20}{3}\ hr$

Speed of wind $v=50\ mph$

Suppose x be the distance between Penthaven and Jackson and u be the speed of plane

So initially

$6.4=\frac{x}{u}+\frac{x}{u}$

$6.4=\frac{2x}{u}$

$x=3.2u \quad \ldots(i)$

When wind is blowing then,

$\Rightarrow \frac{20}{3}=\frac{x}{u+v}+\frac{x}{u-v}$

$\Rightarrow \frac{20}{3}=x[\frac{1}{u+50}+\frac{1}{u-50}]$

$\Rightarrow \frac{20}{3}=x[\frac{2u}{u^2-50^2}]\quad \ldots(ii)$

Substitute the value of x in $(ii)$

$\Rightarrow \frac{20}{3}=\frac{2u[3.2u]}{u^2-50^2}$

$\Rightarrow 10[u^2-50^2]=9.6u^2$

$\Rightarrow 0.4u^2=50^2\times 10$

$\Rightarrow u^2=\frac{50^2\times 10^2}{4}$

$\Rightarrow u=250\ mph$

Thus $x=3.2\times 250=800\ miles$