An airplane flew 3.5 hours with a 40 mph head wind. The return trip with a tail wind of the same speed took 2.5 hours. Find the speed of the plane in still air.

240 mph

220 mph

200 mph

180 mph

Question

An airplane flew 3.5 hours with a 40 mph head wind. The return trip with a tail wind of the same speed took 2.5 hours. Find the speed of the plane in still air. 

 240 mph 

 220 mph 

 200 mph 

 180 mph

whpalmer4 · Answer

$$d = r t$$distance is rate * time
speed of plane in still air is $v$
speed of wind is $v_w$
when the plane flies with the wind (tail wind), it goes $v+v_w$ over the ground (still only $v$ relative to the air, which is moving)
when the plane flies into the wind (head wind), it goes $v - v_w$ over the ground

the distance out and back is the same, so we don't actually need to know it!

$$d = (v+v_w)(2.5) = (v-v_w)(3.5)$$
$$(v+v_w)(2.5) = (v-v_w)(3.5)$$$$2.5 v + 2.5v_w = 3.5 v - 3.5v_w$$Solve for $v$ and plug in the value of $v_w$.

whpalmer4 · Answer

to check your work (essential!), find out how far the plane flies at v-40 for 3.5 hours and compare with how far it flies at v+40 for 2.5 hours.  They should be the same!

whpalmer4 · Answer

what do you get for the answer?

anonymous · Answer

how am I suppose to plug it in? I don't know how to solve that for the answer

whpalmer4 · Answer

Add subtract multiply divide to get v on one side of the equals sign...then you can substitute the value of $v_w = 40$ and get the value of $v$

whpalmer4 · Answer

$$2.5v+2.5v_w = 3.5v - 3.5v_w$$what do you get if you subtract 2.5v from both sides?
$$2.5v - 2.5v + 2.5v_w = 3.5v -2.5v - 3.5v_w$$$$2.5v_w=1.0v-3.5v_w$$Now can you solve that for $v$?

anonymous · Answer

6.0vw=1.0v

whpalmer4 · Answer

yes, so now replace $v_w$ with 40 and you have your answer...

whpalmer4 · Answer

I don't like to replace numbers until all the algebra is done

anonymous · Answer

so 240

whpalmer4 · Answer

yep!  try it out as suggested to make sure it works.

3.5 hours at 200, 2.5 hours at 280