{"id":762,"date":"2010-05-26T17:04:08","date_gmt":"2010-05-27T00:04:08","guid":{"rendered":"https:\/\/c21-wp.phas.ubc.ca\/index.php\/cheetah-chase"},"modified":"2023-03-31T14:17:17","modified_gmt":"2023-03-31T21:17:17","slug":"cheetah-chase","status":"publish","type":"article","link":"https:\/\/c21.phas.ubc.ca\/article\/cheetah-chase\/","title":{"rendered":"PRIVATE: Cheetah Chase"},"content":{"rendered":"
The fastest land mammal, the cheetah, is able to accelerate from a standing start to 96 km\/h in just three seconds[note]Open Learning.\u00a0Studying mammals: meat eaters: Characteristics of the hunters<\/em> (online).\u00a0 http:\/\/openlearn.open.ac.uk\/mod\/oucontent\/view.php?id=398752§ion=1.3.1<\/a>[17 August 2009].[\/note], which corresponds to an acceleration of 8.9 m\/s2<\/sup>. It seems that anything trying to escape would have little chance, right? Cheetahs can only keep up their fastest pace (110 km\/h) for approximately 400 m before their body overheats and their muscles begin to tire and produce lactic acid from exhaustion. In fact, the cheetah’s body temperature can rise to over 41 degrees Celsius, a fatal temperature[note]Blue Lion.\u00a0 Cheetah<\/em> (online).\u00a0 http:\/\/www.bluelion.org\/cheetah.htm<\/a>[17 August 2009][\/note].<\/p>\n Thomson’s gazelles[note]Wikipedia.\u00a0 Thomson’s Gazelle<\/em> (online).\u00a0http:\/\/en.wikipedia.org\/wiki\/Thomson’s_Gazelle <\/a>[17 August 2009].[\/note], their favourite prey, can reach speeds of 70 km\/h. They have an acceleration[note] McNeill Alexander, R. Principles of animal locomotion <\/i>. p.3, Princeton University Press, 2003.[\/note] of approximately 4.5 m\/s2<\/sup>. The gazelle has good endurance and has the additional ability of making sharp turns quickly.<\/p>\n <\/p>\n Figure 1.<\/b> A cheetah and a gazelle[note]Wikipedia.\u00a0 Thomson’s Gazelle<\/em> (online).\u00a0http:\/\/en.wikipedia.org\/wiki\/Thomson’s_Gazelle <\/a>[17 August 2009].[\/note] [note]Wikipedia.\u00a0 Cheetah<\/em> (online).\u00a0http:\/\/en.wikipedia.org\/wiki\/Cheetah <\/a>[17 August 2009][\/note].<\/p>\n What is the maximum distance away from a gazelle for the cheetah to have a chance of catching up to it?<\/p>\n Assumptions:<\/u><\/p>\n Strategy:<\/u><\/p>\n The cheetah would just barely be able to catch the gazelle if it caught it when the cheetah had travelled 400 m at top speed.<\/p>\n Calculations:<\/u><\/p>\n Figure 2.<\/b> The cheetah starts at A, accelerates until B and travels top speed until E. The Gazelle starts at C, accelerates until D and travels top speed until E. The distance the gazelle started away from the cheetah is AC.<\/p>\n<\/div>\n Step 1:<\/u><\/p>\n The cheetah can accelerate from 0 km\/h to 96 km\/h in 3 seconds, this gives an acceleration of ac<\/sub> = 8.9 m\/s2<\/sup>. It has a top speed of vfc<\/sub> = 110 km\/h = 30.6 m\/s. The time to reach its top speed from rest, vic<\/sub> = 0, is<\/p>\n $t_{c1} = \\dfrac{v_{fc} – v_{ic}}{a_c} = 3.4 \\textnormal{ s}$<\/p>\n The time t<\/i>c2<\/sub>to travel at top speed v<\/i>fc<\/sub>, for a distance d<\/i>c2<\/sub> = 400 m is given by<\/p>\n $t_{c2} = \\dfrac{d_{c2}}{v_{fc}} = 13.1 \\textnormal{ s}$<\/p>\n The total time to accelerate to top speed and run 400 m is t<\/i>c<\/sub> = t<\/i>c1<\/sub> + t<\/i>c2<\/sub> = 16.5 s.<\/p>\n Step 2:<\/u><\/p>\n While accelerating for time t<\/i>c1<\/sub>, the cheetah travels a distance of<\/p>\n $d_{c1} = \\dfrac{v_{fc}^2}{2a_c} = 52.5 \\textnormal{ m}$<\/p>\n The cheetah then travels d<\/i>c2<\/sub> = 400 m, so that the total distance travelled by the cheetah is d<\/i>c<\/sub> = 452.5 m.<\/p>\n Step 3:<\/u><\/p>\n The gazelle starts at an initial speed v<\/i>ig<\/sub> = 0 m\/s and accelerates to a top speed of v<\/i>fg<\/sub> = 70 km\/h = 19.4 m\/s in time<\/p>\n $t_{g1} = \\dfrac{v_{gf} – v_{gi}}{a_g} = 4.3 \\textnormal{ s}$<\/p>\n During this initial acceleration stage the gazelle travels a distance<\/p>\n $d_{g1} = \\dfrac{v_{vg}^2}{2a_g} = 42.0 \\textnormal{ m}$<\/p>\n <\/p>\n The gazelle travels at top speed for time<\/p>\n $t_{g2} = t_{tc} – t_{g1} = 12.2 \\textnormal{ s}$<\/p>\n At constant speed the gazelle travelled<\/p>\n $d_{g2} = (v_{gf})(t_{g2}) = (19.4 \\textnormal{ m}\\bullet \\textnormal{s}^{-1})(12.2 \\textnormal{ s}) = 236.7 \\textnormal{ m}$<\/p>\n The gazelle travelled a total distance of d<\/i>g<\/sub> = d<\/i>g1<\/sub> + d<\/i>g2<\/sub> = 278.7 m.<\/p>\n Step 4:<\/u><\/p>\n The cheetah travelled a total distance of d<\/i>c<\/sub> = 452.5 m and the gazelle travelled a total distance of d<\/i>g<\/sub> = 278.7 m. This gives the gazelle a head start of d<\/i>c<\/sub> – d<\/i>g<\/sub> = 174 m before the cheetah is able to catch the gazelle.<\/p>\n Summary:<\/b><\/p>\n Taking into account both acceleration and constant motion we calculated a 174 m distance beyond which the gazelle could not be caught by the cheetah if they both started running at the same time. This assumes linear motion, in real life gazelles use turning maneuvres to escape and the cheetah will stealthily come up to within 10-30 m and then initiate a chase[fn value=5][\/fn].<\/p>\n","protected":false},"author":14,"featured_media":1570,"template":"","tags":[],"date_post_made_public":"2010-05-26","post_authored_by":"Janelle van Dongen","hook":"How far away can a cheetah be from a gazelle and still be guaranteed to catch it?","big_ideas":"<\/p>\n
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