Cheetah Chase

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How far away can a cheetah be from a gazelle and still be guaranteed to catch it?

Big Ideas: 
  • The chasing of a gazelle by a cheetah can be examined using acceleration and constant motion.

The fastest land mammal, the cheetah, is able to accelerate from a standing start to 96 km/h in just three seconds1, 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 aproximately 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 Celcius, a fatal temperature2.

Thomson's gazelles3, their favourite prey, can reach speeds of 70 km/h. They have an acceleration4 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 gazelle3 5.

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, tc, it takes to catch for the cheetah to first accelerate to top speed in time tc1 and then travel 400 m at top speed in time tc2 where tc = tc1 + tc2.
  2. Calculate the total distance travelled by the cheetah dc = dc1 + dc2 where dc1 and dc2 = 400 m are the distances travelled in times tc1 and tc2, respectively.
  3. Calculate how far the gazelle can travel in time tc, dg = dg1 + dg2, where dg1 is the distance travelled by the gazelle when it accelerates to top speed in time tg1 and dg2 is the distance travelled during tg2 = tc - tg1.
  4. Subtract the distance travelled by the gazelle, dg, from the total distance travelled by the cheetah dc. The maximum distance the gazelle can be away and still get caught is dc - dg.

Calculations:

Figure 2. 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.

Step 1:

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

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

The time tc2to travel at top speed vfc, for a distance dc2 = 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 tc = tc1 + tc2 = 16.5 s.

Step 2:

While accelerating for time tc1, the cheetah travels a distance of

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

The cheetah then travels dc2 = 400 m, so that the total distance travelled by the cheetah is dc = 452.5 m.

Step 3:

The gazelle starts at an initial speed vig = 0 m/s and accelerates to a top speed of vfg = 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 dg = dg1 + dg2 = 278.7 m.

Step 4:

The cheetah travelled a total distance of dc = 452.5 m and the gazelle travelled a total distance of dg = 278.7 m. This gives the gazelle a head start of dc - dg = 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 chase5.

Resources
Lecture Notes: 
Cheetah Chase Lecture Notes

Comments

This is great, I am always

This is great, I am always looking for good biology relevant examples for life science students. There is one typo I saw. In step 1 you have 100 km/hr instead of 110 km/hr, which is what all the math uses and is in the problem statement.

Thanks for letting us know. I

Thanks for letting us know. I have corrected the typo now.

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