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15 Oct 2006, 08:31
Kudos
Bookmarks
Time taken by leopard = time taken by the Deer
(Distance/Rate)L = (Distance/Rate)D
Let the distance travelled by the Deer be X
200+X/12 = X/10
solving for X, X = 1000m or 1km
Hope this helps....
(Distance/Rate)L = (Distance/Rate)D
Let the distance travelled by the Deer be X
200+X/12 = X/10
solving for X, X = 1000m or 1km
Hope this helps....
15 Oct 2006, 09:38
Kudos
Bookmarks
Sorry, I should have been a bit clearer about which guy. I understand the basic principle. I do not understand the differential speed approach.
Artemov says,
I don't understand why we are using 0.1hr to arrive at the distance traveled by the deer before it is caught when, clearly, the lion took that time to arrive at the deer's initial position. Shouldn't the time we use be greater?
Artemov says,
Quote:
I don't understand why we are using 0.1hr to arrive at the distance traveled by the deer before it is caught when, clearly, the lion took that time to arrive at the deer's initial position. Shouldn't the time we use be greater?
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