Problem using this formula both ways

[dsryan]

Okay, so I've messed about with this for a while now and I've found that the following formula works on how to get the speed.

A car has covered 27 miles in 120 minutes. What is the speed it is traveling at?

I add three zeros to the 27, making it 27,000.

120/27000 = 225. I then put a decimal point at the start of the three digit number (if it were a 4 digit number, for example; 2225, I'd make it 2.225) but for this particular sum the number is 0.225.

0.225 x 60 = 13.500mph

I have found this method works, and its very easy!Now, I'm trying to use a similar method to use in order to obtain distances and speed.
For example, could someone answer the following setting it out the same way I did with the mph formula?

A car has been traveling 42 mph for 73 minutes, how far (in miles) has it travelled?

A car has traveled 52 miles in 39 minutes, what is the car's average speed?

 

[MarkFL]

Personally, I find it much easier to write:

\(\displaystyle \overline{v}=\frac{d}{t}=\frac{27\text{ mi}}{120\text{ min}}\cdot\frac{60\text{ min}}{1\text{ hr}}=\frac{27}{2}\text{ mph}=13.5\text{ mph}\)

To me it is much more obvious what is going on. For example, suppose we wish to convert a speed given in mph to m/s. We could simply write:

\(\displaystyle v\text{ mph}=v\frac{\text{mi}}{\text{hr}}\cdot\frac{127\text{ cm}}{50\text{ in}}\cdot\frac{12\text{ in}}{1\text{ ft}}\cdot\frac{5280\text{ ft}}{1\text{ mi}}\cdot\frac{1\text{ m}}{100\text{ cm}}\cdot\frac{1\text{ hr}}{3600\text{ s}}=\frac{1397}{3125}v\frac{\text{m}}{\text{s}}\)

I think you will find this method will serve you better in all types of unit conversions.