What is the speed of advance of the glacier in m/s?

Question

The end of Hubbard Glacier in Alaska advances by an average of 105 feet per year.  What is the speed of advance of the glacier in m/s?

Answer 1

1 foot = .3048 meters

105 ft = 105(.3048) = 32.004 m

 

Contrary to popular belief there are 365.25 days in a year.

That is the reason for adding a day to the calendar every

four years (leap year)

 

# seconds in a year;

 

(365.25 days)(24 hrs/day)(60 min/hr)(60 sec/min)

= 31,557,600 sec/yr

 

 m/sec = 32.004/31557600 = 1.0141455624 x 10^-6 m/sec

 

Round to 1.01 x 10^-6 m/sec

 

 

Answer 2

This is a problem that is all about conversions. Typing them in properly is tough, but I will try my best.

 

105 feet/year. Feet is in the numerator of the units, and we want meters instead. This means we should cancel them out - we should put feet in the denominator of our conversion. 

 

1 ft = 0.3048 m

 

105 (ft/yr) * (0.3048 m / 1 ft) = 32.004 m / year

 

Now we want to convert years to seconds. This is a bunch of conversions (years to days, then days to hours, then hours to minutes, then minutes to seconds)

 

32.004 (m/yr) * (1 yr / 365 days) * (1 day / 24 hours) * (1 hour / 60 min) * (1 minute / 60 seconds)

 

All those units cancel out and we get m/s. Let me re-write the conversion without the units so that it is cleaner

 

32.004 * (1/365) * (1/24) * (1/60) * (1/60) = 0.00000101484 m/s

 

This is a case where scientific notation shines - it really tough to see how many 0's there are between the decimal point and the first number.

 

1.01x10^-6 m/s

 

I hope this helps. Please ask any questions if you have them.