In a 10 mile race, Janet covered the first 2 miles at a constant rate. She then sped up and rode her bike the last 8 miles at a rate that was 0.5 miles per minute faster. Janet's overall time would have been 2 minutes faster had she ridden her bike the whole race at the faster pace. What was Janet's average speed (in miles per minute) for the whole race?

Guest Apr 7, 2020

#1**+1 **

Let x be the speed for the first two miles and x+.5 be the speed for the last 8 miles. Since D=rt, \(2\over x \)+\(8\over x+.5\)is the time it took Janet to do the 10 mile race. If she took her bike the whole way, her time would be \(10\over x+.5\). Since that was two minutes faster than her real time, \(2\over x \)+\(8\over x+.5\)-\(10\over x+.5\)=2. Solving for s you get s=.5. The question asks for the average speed for the whole race so, \(2\over .5\)+\(8\over .5+.5\)=\(10\over a\) where a is the average speed. A= **\(5\over 6\) miles per minute**.

Hope it helps!

HELPMEEEEEEEEEEEEE Apr 7, 2020

#2**+1 **

2miles / f mile/m + 8 miles / (f+.5) mile/min = 2min + 10 miles / (f+.5) mile /min

solve for 'f'

2/f + 8(f + .5) = 2 + 10/(f + .5)

2/f = 2 + 2/(f+.5)

2(f+.5) = 2(f+.5)f + 2f

2f+1 = 2f^2 +f + 2f

2f^2 +f - 1 = 0 f = .5 then faster speed = 1 mile /min

2miles / .5 mile/min + 8 miles/ 1 mile/min = 12 minutes for 10 miles

10 miles/12 min = 5/6 mile / min average (Note: this is 50 miles per hour average....on a bike ! Was it a MOTORbike ??)

ElectricPavlov Apr 7, 2020