Chapter 2: Q. 53 (page 62)

A cheetah spots a Thomson’s gazelle, its preferred prey, and
leaps into action, quickly accelerating to its top speed of 30 m/s, the highest of any land animal. However, a cheetah can maintain this extreme speed for only 15 s before having to let up. The cheetah is 170 m from the gazelle as it reaches top speed, and the gazelle sees the cheetah at just this instant. With negligible reaction time, the gazelle heads directly away from the cheetah, accelerating at $4.6 m / s^{2}$ for 5.0 s, then running at constant speed. Does the gazelle escape? If so, by what distance is the gazelle in front when the cheetah gives up?

Expert verified

The cheetah will give up at $7.5 m$ distance in front of the gazelle.

Step by step solution

The extreme speed possessed by the cheetah, $V_{C} = 30 m / s .$

The cheetah can maintain the top speed for a duration of, $t = 15 s .$

Initial distance of the gazelle from the cheetah is $170 m .$

Acceleration of the gazelle, $a = 4.6 m / s^{2}$ for a time of $t' = 5 s .$

The total distance that the cheetah covered in time $t$ is,

$s = V_{C} t s = 30 m / s \times 15 s s = 450 m$

Therefore, the cheetah needs to cover $(450 - 170) m = 280 m$ to escape.

In time $t' = 5 s$ , the gazelle escapes at a distance,

$x = 0 + \frac{1}{2} a t'^{2} x = \frac{1}{2} \times 4.6 \times 5^{2} x = 57.5 m$

The constant speed of the gazelle is,

role="math" localid="1648460424964"

v_{g} = a t v_{g} = 4.6 m / s^{2} \times 5 s v_{g} = 23 m / s

Rest of the distance travelled by the gazelle with constant speed in $(15 - 5) s = 10 s$ is,

$d = v_{g} t d = (23 \times 10) m d = 230 m$

In $15 s$ the gazelle covers a distance of $(230 + 57.5) m = 287.5 m$ .

The gazelle will led by $(287.5 - 280) m = 7.5 m$ distance compare to the cheetah.

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