In a 21-meter race between a tortoise and a hare, the tortoise leaves 9 minutes before the hare. The hare, by running at an average speed of 0.5 meters per hour faster than the tortoise, crosses the finish line 3 minutes before the tortoise. What are the average speeds of the tortoise and the hare?
My thinking is this:
x(y+3/20)=21
(x+.5)(z)=21
My thinking is this:
x(y+3/20)=21
(x+.5)(z)=21
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In 9 minutes, the hare has went (t + 0.5)(0.15) meters where t is the speed of the tortoise. The tortoise crosses the finish line after 21/t hours and the hare crosses the finish line in (21/t - 1/20) hours.
(t + 0.5)(0.15) + (21/t - 1/5)(t + 0.5) = 21
Now try solving for t:
0.15t + 0.075 + 21 + 11.5/t - t/5 - 0.1 = 21
11.5/t - t/20 = 0.025
20(11.5) - t² = (20t)(0.025)
t² + 0.5t - 230 = 0
2t² + t - 460 = 0
t = (-1 +/- √(1 - 4(2)(-460))/4 = 14.92 m/hr
(t + 0.5)(0.15) + (21/t - 1/5)(t + 0.5) = 21
Now try solving for t:
0.15t + 0.075 + 21 + 11.5/t - t/5 - 0.1 = 21
11.5/t - t/20 = 0.025
20(11.5) - t² = (20t)(0.025)
t² + 0.5t - 230 = 0
2t² + t - 460 = 0
t = (-1 +/- √(1 - 4(2)(-460))/4 = 14.92 m/hr