A few days ago
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
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!
0
A few days ago
How much “over 11 miles”?
0
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