Question

Which of the following would be a correct use of dimensional analysis to convert 45 mph to feet per sec? a. $\frac{45 \: miles}{hour} \times \frac{1mile}{5280ft} \times \frac{1hour}{60min} \times \frac{1min}{60sec}$b. $\frac{45miles}{hour} \times \frac{5280ft}{1mile} \times \frac{1hour}{60min} \times \frac{1min}{60sec}$c. $\frac{45miles}{1hour} \times \frac{1ft}{5280miles} \times \frac{1hour}{60min} \times \frac{1min}{60sec}$

Answer 1

We have the following conversion regarding that:

1 mile = 5280.02 feet

1 hour = 3600 sec

Then, to convert 45 mph to feet per sec, we have:

$45\frac{m}{h}\cdot5280.02\frac{f}{m}\cdot\frac{1h}{3600s}\Rightarrow\frac{45\cdot5280.02}{3600}ft/\sec$

The units m/m = 1, h/h =1, and then the units remaining were feet / sec.

Then, we have:

$\frac{45\cdot5280.02}{3600}=66.00025ft/\sec$

Then, the two conversions are:

• 45 m/h * 5280.0 feet/mile, and

,

• 1 h/ 3600sec = (1 hour/60min) * (1 min/60sec).