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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?

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?
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"#.


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