# Mathematics - Other

Years after the famous foot race, the hare challenged the tortoise again on bicycles. The hare was able to pedal at a rate of 15 mph. while the tortoise could pedal only 12 mph. Knowing he was slower, the tortoise makes up a lame excuse for being given a 27 mi. head start in the 165 mi. race.

1. WHo won the tortoise and hard race? By how much?

2. How far into the race did the hare overtake the tortoise?

3. What is the largest head start the tortoise could have given without being able to beat the hare?

4. Chipmunks are fast on bikes. Suppose a chipmunk entered the race with an 84 mi. head start ahnd traveled at a rate of 9 mph. What is an equation for the chipmunk. When would the hard finally catch up.

(assuming the race does not end at 165 mi.)

A confused pig also followed the race. The pig started at the finsih line at a rate of 15 mph and rode his bike backwards. Where and when did he meet the hare on the road?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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Years after the famous foot race, the hare challenged the tortoise again on bicycles. The hare was able to pedal at a rate of 15 mph. while the tortoise could pedal only 12 mph. Knowing he was slower, the tortoise makes up a lame excuse for being given a 27 mi. head start in the 165 mi. race.

1. WHo won the tortoise and hard race? By how much?

2. How far into the race did the hare overtake the tortoise?

3. What is the largest head start the tortoise could have given without being able to beat the hare?

4. Chipmunks are fast on bikes. Suppose a chipmunk entered the race with an 84 mi. head start ahnd traveled at a rate of 9 mph. What is an equation for the chipmunk. When would the hard finally catch up.

(assuming the race does not end at 165 mi.)

A confused pig also followed the race. The pig started at the finsih line at a rate of 15 mph and rode his bike backwards. Where and when did he meet the hare on the road?

Distance = 165 miles, hare's speed = 15 mph, time taken by hare to finish the race = 165/15 = 11 hrs.

For tortoise, d = 165 - 27 = 138 miles, speed of tortoise = 12 mph, time taken by tortoise to finish the race = 138/12 = 11.5 hrs.

(a) Thus, the hare won the race by 0.5 hrs or 30 mins.

(b) Let the hare overtake the tortoise after covering a distance x. Time taken by it to cover x miles = x/15 hrs. The distance covered by the tortoise in the same time = x - 27 and time taken is (x - 27)/12. Equating the times, x/15 = (x - 27)/12 ï‚® 12x = 15(x - 27) ï‚® x = 135 miles.

(c) Let the head start given to the tortoise be y. Then, tortoise's time =

(165 - y)/12 hrs and hare's time = 11 hrs.

Tortoise's time just > hare's time ï‚® (165 - y)/12 > 11 ï‚® 165 - y > 132 ï‚®

y - 165 < -132 ï‚® y < 165 - 132 ï‚® y < 33 miles.

(d) Let the hare catch up with the chipmunk after covering D miles.

For the chipmunk, distance = D - 84 miles. It covers this distance in a time of

(D - 84)/9 hrs. The hare must cover distance D in the same time ï‚®

D/15 = (D - 84)/9 ï‚® 15(D - 84) = 9D ï‚® D = 210 miles.

(e) Let the pig meet the hare after time t hrs. Distance covered by the pig

in t hrs = 15t miles. The distance covered by the hare in t hrs = 15t miles.

Total distance = 165 miles ï‚® 15t + 15t = 165 ï‚® t = 5.5 hrs.

Distance of meeting = 15 x 5.5 = 82.5 miles (Middle point of the race distance).

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