# If one mile is approximately 1606 m, 55 mph is (A) 15 m/s (B) 81 m/s (C) 25 m/s (D) 123 m/s

## Question:

If one mile is approximately {eq}1606 \, \mathrm{m} {/eq}, {eq}55 \, \mathrm{mph} {/eq} is

(A) {eq}15 \, \mathrm{m/s} {/eq}

(B) {eq}81 \, \mathrm{m/s} {/eq}

(C) {eq}25 \, \mathrm{m/s} {/eq}

(D) {eq}123 \, \mathrm{m/s} {/eq}

## Unit Conversion:

We can convert a quantity into different units (such as meters to miles or Newtons to pounds) as long as they are of the same dimension. Unit analysis is a way of changing from one unit to another.

To convert into different units, we need a conversion factor: a way of writing the same quantity in two different units. For example, there are 60 seconds in one minute. So, consider the fraction:

{eq}\dfrac{60 \ s}{1 \ min} {/eq}

The numerator and the denominator of this fraction both represent the same thing. Any quantity divided by itself is one, so the fraction above is equal to one. This means that we can multiply it by any quantity, without changing the value of that quantity.

So, let's try multiplying this conversion factor by a number of minutes:

{eq}4 \ min * \dfrac{60 \ s}{1 \ min} {/eq}

The unit of minutes in the denominator cancels out with the unit of minutes sitting in front. We can rewrite this expression as:

{eq}4 * \dfrac{60 \ s}{1}\\ 240 \ s {/eq}

We have successfully converted 4 minutes into seconds.

We are asked to convert a speed from mph to meters per second. Let's write out this speed explicitly.

{eq}v = \dfrac{55 \ mi}{1 \...

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