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?
+934 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!
+36916 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 ??)
