A few days ago
SCoof

A good math problem about the hare and the tortoise.?

In a race the tortoise is given a head start of over 11 miles over the hare. The hare runs at 6 mph and the tortoise travels at 4 mph. When and where does the hare catch up with the tortoise? What if the hare was going at 6 km/h?

if you answer show work thanks

Top 2 Answers
A few days ago
Fahd

Favorite Answer

when both of ’em meet, their distance from a fixed point should be same. Keeping that in mind….

let x = time in hours after the race when they meet.

6x = distance travelled by the hare

4x+11= distance travelled by the tortoise

hence, when they meet

6x = 4x+11 (assuming a head start of 11 miles)

or, 2x = 11

or, x = 11/2 h = 5.5 hours

where?

At 6 X 5.5 = 33 miles.

6 km = 3.73 miles

if the hare travelled at 6km/h… a little tricky..ha!

3.73x = 4x+11

the solution is negative, hence they would never meet at their designated speeds which remain constant.

You would know if someone copied me!

Cheers!

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A few days ago
daleabono
How much “over 11 miles”?
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