1. ## Motion Problem

Just need clarification...

Here's the problem.

A plane flew from a southern city to a northern city, a distance of 900 miles, against a 30mph headwind. On the return flight, the 30mph wind became the tailwind. The plane was in the air for a total of 5 and a half hours. What would the plane's average speed have been without the wind?

Isn't it the plane's average speed would still be the same?

so?
V=d/t=900(2)/5.5
V=327.2727mph

However, I'm not really sure of this...

Would appreciate the Help.

2. Hello mcroldan08
Originally Posted by mcroldan08
Just need clarification...

Here's the problem.

A plane flew from a southern city to a northern city, a distance of 900 miles, against a 30mph headwind. On the return flight, the 30mph wind became the tailwind. The plane was in the air for a total of 5 and a half hours. What would the plane's average speed have been without the wind?

Isn't it the plane's average speed would still be the same?
No. Just suppose, for a minute, that the plane is a bird that can only fly at 30 mph. Then against the headwind it won't make any progress at all, and would take an infinite time to get there. So its average speed would be zero. Whereas without the wind, it would obviously be 30 mph.

You have to suppose that in still air the plane flies at $\displaystyle x$ mph. Then against the wind it flies at $\displaystyle (x-30)$ mph, and with the wind it flies at $\displaystyle (x+30)$ mph.

Then use the formula
$\displaystyle \text{time}=\frac{\text{distance}}{\text{speed}}$
to write down the times taken for each leg of the journey; add them together and equate the answer to 5.5 hours. I'll start you off...
$\displaystyle \frac{900}{x-30}+ ... = ...$
To solve this equation, multiply throughout by $\displaystyle 2(x-30)(x+30)$ to get rid of fractions. You'll then get a quadratic equation with only one positive root. There's your answer.

Can you complete this now?