a.) Lance Armstrong captured the lead with an impressive climbto the finish line

Question

a.) Lance Armstrong captured the lead with an impressive climbto the finish line, elevation 1560 m. Armstrong made the 13.5 kmclimb in 36 min. What is his average speed? b.) If the altitude at the base of the climb was 493.5 m, whatwas his average vertical velocity? a.) Lance Armstrong captured the lead with an impressive climbto the finish line, elevation 1560 m. Armstrong made the 13.5 kmclimb in 36 min. What is his average speed? b.) If the altitude at the base of the climb was 493.5 m, whatwas his average vertical velocity?

Explanation / Answer

So, in order to calculate the average speed, you need to take thedistance traveled (13.5 km) and divide it by the amount of time ittook him to travel that distance (36 min) so, average speed = 13.5/36 (and the units will bekm/min) For part b, the total vertical distance climbed is equal to theheight of elevation of the finish line subtracted by the height ofelevation of the base (1560 - 493.5) h = 1560-493.6= 1066.5 meters. from here it's just like part a. For average VERTICAL velocity, youtake the total VERTICAL distance traveled divided by the time ittook to travel it. average speed = h/36 = 1066.5/36   (these units will bein meters/min) Hope this helps.

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