PRIVATE: Cheetah Chase

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. Studying mammals: meat eaters: Characteristics of the hunters (online).  http://openlearn.open.ac.uk/mod/oucontent/view.php?id=398752&section=1.3.1[17 August 2009].[/note], which corresponds to an acceleration of 8.9 m/s². 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.  Cheetah (online).  http://www.bluelion.org/cheetah.htm[17 August 2009][/note].

Thomson’s gazelles[note]Wikipedia.  Thomson’s Gazelle (online). http://en.wikipedia.org/wiki/Thomson’s_Gazelle [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 . p.3, Princeton University Press, 2003.[/note] of approximately 4.5 m/s². The gazelle has good endurance and has the additional ability of making sharp turns quickly.

 

Figure 1. A cheetah and a gazelle[note]Wikipedia.  Thomson’s Gazelle (online). http://en.wikipedia.org/wiki/Thomson’s_Gazelle [17 August 2009].[/note] [note]Wikipedia.  Cheetah (online). http://en.wikipedia.org/wiki/Cheetah [17 August 2009][/note].

What is the maximum distance away from a gazelle for the cheetah to have a chance of catching up to it?

Assumptions:

• The cheetah and gazelle start from rest and start accelerating at the same time.
• The rate of acceleration from rest to top speed is constant.
• The chase only takes place in one dimension, i.e. the gazelle only runs straight.

Strategy:

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.

1. Calculate the total time, t_c, it takes to catch for the cheetah to first accelerate to top speed in time t_c1 and then travel 400 m at top speed in time t_c2 where t_c = t_c1 + t_c2.
2. Calculate the total distance travelled by the cheetah d_c = d_c1 + d_c2 where d_c1 and d_c2 = 400 m are the distances travelled in times t_c1 and t_c2, respectively.
3. Calculate how far the gazelle can travel in time t_c, d_g = d_g1 + d_g2, where d_g1 is the distance travelled by the gazelle when it accelerates to top speed in time t_g1 and d_g2 is the distance travelled during t_g2 = t_c – t_g1.
4. Subtract the distance travelled by the gazelle, d_g, from the total distance travelled by the cheetah d_c. The maximum distance the gazelle can be away and still get caught is d_c – d_g.

Calculations:

Step 1:

The cheetah can accelerate from 0 km/h to 96 km/h in 3 seconds, this gives an acceleration of a_c = 8.9 m/s². It has a top speed of v_fc = 110 km/h = 30.6 m/s. The time to reach its top speed from rest, v_ic = 0, is

$t_{c1} = \dfrac{v_{fc} – v_{ic}}{a_c} = 3.4 \textnormal{ s}$

The time t_c2to travel at top speed v_fc, for a distance d_c2 = 400 m is given by

$t_{c2} = \dfrac{d_{c2}}{v_{fc}} = 13.1 \textnormal{ s}$

The total time to accelerate to top speed and run 400 m is t_c = t_c1 + t_c2 = 16.5 s.

Step 2:

While accelerating for time t_c1, the cheetah travels a distance of

$d_{c1} = \dfrac{v_{fc}^2}{2a_c} = 52.5 \textnormal{ m}$

The cheetah then travels d_c2 = 400 m, so that the total distance travelled by the cheetah is d_c = 452.5 m.

Step 3:

The gazelle starts at an initial speed v_ig = 0 m/s and accelerates to a top speed of v_fg = 70 km/h = 19.4 m/s in time

$t_{g1} = \dfrac{v_{gf} – v_{gi}}{a_g} = 4.3 \textnormal{ s}$

During this initial acceleration stage the gazelle travels a distance

$d_{g1} = \dfrac{v_{vg}^2}{2a_g} = 42.0 \textnormal{ m}$

 

The gazelle travels at top speed for time

$t_{g2} = t_{tc} – t_{g1} = 12.2 \textnormal{ s}$

At constant speed the gazelle travelled

$d_{g2} = (v_{gf})(t_{g2}) = (19.4 \textnormal{ m}\bullet \textnormal{s}^{-1})(12.2 \textnormal{ s}) = 236.7 \textnormal{ m}$

The gazelle travelled a total distance of d_g = d_g1 + d_g2 = 278.7 m.

Step 4:

The cheetah travelled a total distance of d_c = 452.5 m and the gazelle travelled a total distance of d_g = 278.7 m. This gives the gazelle a head start of d_c – d_g = 174 m before the cheetah is able to catch the gazelle.

Summary:

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].