## Question

###### The diameter of the solar system is approximately: 7,500,000,000 miles. How long would it take to drive this distance if traveling 60 mph?

## Answers

14.26 millenia, or 125,000,000 hours.

## Explanation:

When we're dealing with numbers this large, it can help to convert them to scientific notation before performing calculations with them.

#7,500,000,000# is#7.5times10^9# in scientific notation, and#60# is simply#6times10# . To find the time it would take to travel#7.5times10^9# miles, we divide it by the rate of#6times10# mph, obtaining:

#(7.5times10^9 \ "mi")/(6times10\ "mi/hr")=7.5/6times10^8\ "hr"# We find that

#7.5/6# gives us#1.25# , leaving us with#1.25times10^8# or#125,000,000# hours. We could stop there, but to get a feel for just how long this is, it'd help to convert it to a more sensible timescale.Let's first convert those hours to years. To make the conversion, we'll use the unit rates of

#(1\ "day")/(24\ "hr")# and#(1\ "yr")/(365\ "days")# :Hours to days:

#1.25times10^8\ "hr"*(1\ "day")/(24\ "hr")=#

#=(1.25times10^8\ cancel("hr"))/(24\ cancel("hr"))*1\ "day"#

#=1.25/24times10^8\ "days"# Days to years:

#1.25/24times10^8\ "days"*(1\ "yr")/(365\ "days")=#

#=1.25/24times10^8\ cancel("days")*1/(365\ cancel("days"))*1\ "yr"#

#=1.25/(24*365)times10^8\ "yr"# We can rewrite

#24# and#365# in scientific notation as#2.4times10# and#3.65times10^2# , obtaining

#1.25/(2.4times10*3.65times10^2)times10^8\ "yr"=#

#= 1.25/(2.4*3.65)times10^8/10^3\ "yr"#

#=1.25/8.76times10^5\ "yr"# Finally, we can use the unit rate of

#(1\ "millennium")/(1000\ "yr")# (or equivalently,#(1\ "millennium")/(10^3\ "yr")# ) to get our answer in millennia:

#1.25/8.76times10^5\ "yr"*(1\ "millennium")/(10^3\ "yr")=#

#=1.25/8.76times10^5\ cancel("yr")*1/(10^3\ cancel("yr"))*1\ "millennium"#

#=1.25/8.76times10^5/10^3\ "millennia"#

#=1.25/8.76times10^2\ "millennia"#

#1.25/8.76approx0.1426# , so our answer in millennia is approximately#0.1426times10^2=14.26\ "millennia"# .