Thomson\'s gazelle: Acceleration: from 0 to 90 km/hr in about 18 seconds. Max sp

Question

Thomson's gazelle:

Acceleration: from 0 to 90 km/hr in about 18 seconds.

Max speed: 90 km/hr

Stamina: Can maintain its maximum speed for long periods of time (compared to the cheetah).

Cheetah:

Acceleration: from 0 to 120 km/hr in 3 seconds

Max speed: 120 km/hr but can only maintain it for about 30 seconds.

After its initial high speed sprint, it collapses due to heat exhaustion -- its temperature may rise as high as 105o F -- and itself becomes vulnerable to larger predators.

A. Let's begin by considering a very simple model of predator-prey interaction. Let's assume that the predator creeps up on a resting prey to a distance, d, and catches the prey during the period when they are both accelerating. At time t = 0, the prey sees the predator and both take off, accelerating as fast as they can. Throughout this exercise, you can assume the animals have constant acceleration until reaching maximum speed.

A.1 The cheetah only accelerates for about 3 seconds. If they both start accelerating at the same time, what will the gazelle's and cheetah's average velocity be for that 3-second time interval?

A.2 How far will they each have traveled in that time interval? How far away does the gazelle have to start in order for the cheetah not to catch it during the cheetah's acceleration phase?

B. Now let's refine the model by including the cheetah's high speed sprint as well as its acceleration. We notice that the cheetah can run faster than the gazelle, but only for 30 seconds. If they are both running straight, how far away must the gazelle be from the cheetah in order that the cheetah not catch it before the cheetah collapses from exhaustion?

C. Find a symbolic representation of the maximum distance, d, for which a general predator can catch its prey if:

The predator can accelerate at a rate of a1 for a time t1 and can run at that final speed for a time T1. After that time, it must stop running.

The prey can accelerate at a rate of a2 for a time t2 and can run at that final speed for a long time -- at least for a lot longer than the predator can run.

Photo by Jason Bechtel
CC permissio

Thomson's gazelle:

Acceleration: from 0 to 90 km/hr in about 18 seconds.

Max speed: 90 km/hr

Stamina: Can maintain its maximum speed for long periods of time (compared to the cheetah).

Cheetah:

Acceleration: from 0 to 120 km/hr in 3 seconds

Max speed: 120 km/hr but can only maintain it for about 30 seconds.

After its initial high speed sprint, it collapses due to heat exhaustion -- its temperature may rise as high as 105o F -- and itself becomes vulnerable to larger predators.

A. Let's begin by considering a very simple model of predator-prey interaction. Let's assume that the predator creeps up on a resting prey to a distance, d, and catches the prey during the period when they are both accelerating. At time t = 0, the prey sees the predator and both take off, accelerating as fast as they can. Throughout this exercise, you can assume the animals have constant acceleration until reaching maximum speed.

A.1 The cheetah only accelerates for about 3 seconds. If they both start accelerating at the same time, what will the gazelle's and cheetah's average velocity be for that 3-second time interval?

A.2 How far will they each have traveled in that time interval? How far away does the gazelle have to start in order for the cheetah not to catch it during the cheetah's acceleration phase?

B. Now let's refine the model by including the cheetah's high speed sprint as well as its acceleration. We notice that the cheetah can run faster than the gazelle, but only for 30 seconds. If they are both running straight, how far away must the gazelle be from the cheetah in order that the cheetah not catch it before the cheetah collapses from exhaustion?

C. Find a symbolic representation of the maximum distance, d, for which a general predator can catch its prey if:

The predator can accelerate at a rate of a1 for a time t1 and can run at that final speed for a time T1. After that time, it must stop running.

The prey can accelerate at a rate of a2 for a time t2 and can run at that final speed for a long time -- at least for a lot longer than the predator can run.

Photo by Jason Bechtel
CC permissio

Explanation / Answer

gazelle:

Acceleration: from 0 to 90 km/hr in about 18 seconds.

Max speed: 90 km/hr = 25 m/s

a = 25/18 m/s2

Cheetah:

Acceleration: from 0 to 120 km/hr in 3 seconds

Max speed: 120 km/hr = 33.333 m/s but can only maintain it for about 30 seconds.

a= 33.333/3 m/s2

A) 3 second period

Average speed gazelle = (0+3*25/18) / 2 = 2.083 m/s

Average speed of cheetah = (0+33.333)/2 = 16.667 m/s

Distance traveled by gazelle = 0.5*(25/18)*32 = 6.25 m

Distance traveled by chetah = 0.5 * 11.111*32 = 50 m

So gazelle has to start 50-6.25 = 43.75 m ahead to avoid getting caught during acceleration of cheetah

B) Distance covered by chetah in 30 s of constant speed = 30*33.33 = 1000 m

Total distance covered by chetah before exhaustion = 1000 + 50 =1050 m

Distance covered by gazelle in 33 second = 0.5*(25/18)*182 + 25*15 = 600 m

So gazelle has to start 1050-600 = 450 m ahead to avoid getting caught during acceleration of cheetah

C) Distance covered by predator in time t1+T1 = 0.5*a1*t12 + a1t1T1

Distance covered by prey in the same time = 0.5*a2*t22 + a2t2(T1+t1-t2)

Maximum distance seperation for which predator can catch its prey = 0.5*a1*t12 + a1t1T1 - 0.5*a2*t22 - a2t2(T1+t1-t2)

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