Converting Rates

ImpossibleDreams

New member
Joined
Oct 18, 2011
Messages
1
Hello. I am having problems with this word problem (no pun intended). Here is the problem:

A model airplane flies 18 feet in 2 seconds. What is the airplane's speed in miles per hour? Round your answer to the nearest hundredth.

This is my work so far:

18 feet times 5280
-------- --------
2 seconds 1 mile

Any problems with my work? I would also like to get help as soon as possible, this assignment is due tomorrow. Thank you!
 
Hi
I don't really get your work but this is how I would do it,

18 feet in 2 seconds means that the plan goes 9feet/second then there are 3600seconds/hour and 1Kilometer/3 280.84Feet we want to find kilometers/hour so we can start crossing out units.

9feet/second* 3600seconds/hour the seconds cancel and we are left with
32 400 feet/hour.

now we need to get rid of the feet so 32 400 feet/hour * 1K/3 280.84Ft= 9.88 kilometers/hour

you could also Google it :) "9 feet per second to kilometer per hour"
 
Last edited:
Hello, ImpossibleDreams!

A model airplane flies 18 feet in 2 seconds.
What is the airplane's speed in miles per hour?
Round your answer to the nearest hundredth.

I was taught this trick in a Physics class . . . very handy!

We are given: .\(\displaystyle \dfrac{18\text{ ft}}{2\text{ sec}}\)

We want to convert it to: .\(\displaystyle \dfrac{\text{miles}}{\text{hour}}\)



Do we know the conversion between feet and miles?
. . Yes . . . \(\displaystyle 5280\text{ feet} \,=\,1\text{ mile}\)

Form a fraction: .\(\displaystyle \dfrac{5280\text{ ft}}{1\text{ mi}}\:\text{ or }\:\dfrac{1\text{ mi}}{5280\text{ ft}}\)


Do we know the conversion between seconds and hours?
. . Yes . . . \(\displaystyle 1\text{ hour} \:=\:60\text{ minutes} \:=\:3600\text{ seconds} \)

Form a fraction: .\(\displaystyle \dfrac{1\text{ hr}}{3600\text{ sec}}\:\text{ or }\:\dfrac{3600\text{sec}}{1\text{ hr}} \)


Multiply our given fraction by the appropriate conversion fractions
. . so that the units will "cancel out".

We use: .\(\displaystyle \dfrac{18\:\text{ft}}{2\:\text{sec}} \times \dfrac{1\:\text{mi}}{5280\:\text{ft}} \times \dfrac{3600\:\text{sec}}{1\;\text{hr}} \)

And we have: .\(\displaystyle \dfrac{18\:\rlap{/\!/}\text{ft}}{2\:\rlap{///}\text{sec}} \times \dfrac{1\:\text{mi}}{5280\:\rlap{/\!/}\text{ft}} \times \dfrac{3600\:\rlap{///}\text{sec}}{1\;\text{hr}} \;=\;\dfrac{18\times3600\text{ mi}}{2\times 5280\text{ hr}}\)

. . . . . . . . \(\displaystyle \dfrac{135\text{mi}}{22\text{ hr}} \;=\;6\frac{3}{22}\text{ mph.}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Long ago, I memorized: .\(\displaystyle 60\text{ mi/hr} \:=\:88\text{ ft/sec}\)

And this gives me yet another opportunity to be the life of the party.


Since .\(\displaystyle \dfrac{88}{60} \:\approx\:\dfrac{90}{60} \:=\:1\frac{1}{2}\), .I can approximate mi-per-hr to ft-per-sec
. . by multiplying by one-and-a-half.

For example, \(\displaystyle 80\text{ mph} \;\approx\;80 + \text{(half of 80}) \;=\;120\text{ ft/sec}\)
 
Hello, Denis!

I am not disagreeing with you . . . I too prefer Simplicity.


The method I showed was one that saved me a lot of grief
. . from my undergraduate years to the present.

I knew that: .\(\displaystyle 1\text{ inch} \:\approx\:2.54\text{ cm}\)
But when I was converting from one unit to the other,
. . the question always arose: Do I multiply or divide by 2.54 ?
And it always took me a few moments to reason it out.

But now I can convert angstroms-per-nanosecond to fathoms-per-fortnight.
. . No problem!
 
Top