From rest, the cheetah can accelerate at 8.8 m/s2 and reach a top speed of 30 m/s (108 km/h)! It can maintain this maximum speed over a distance of about 400 meters before it needs to stop. On the other hand, the Thomson’s gazelle has a top speed of 70 km/h, which is less than the cheetah’s, but it can maintain this top speed for a while as well. From rest, the gazelle can accelerate at 4.5 m/s2 to reach its top speed. When a cheetah goes after a gazelle, success or failure is a simple matter of kinematics: you will determine if the cheetah’s high speed is enough to allow it to reach its prey before it runs out of steam (or time). You will apply basic kinematics and simple assumptions to determine how a chase can play out. Q2. In the elapsed time that the cheetah started and must stop, what distance can the gazelle cover?   (again clearly show your calculations)     Why is the 70 multiplied by 5/18?

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From rest, the cheetah can accelerate at 8.8 m/s2 and reach a top speed of 30 m/s (108 km/h)! It can maintain this maximum speed over a distance of about 400 meters before it needs to stop.
On the other hand, the Thomson’s gazelle has a top speed of 70 km/h, which is less than the cheetah’s, but it can maintain this top speed for a while as well. From rest, the gazelle can accelerate at 4.5 m/s2 to reach its top speed.

When a cheetah goes after a gazelle, success or failure is a simple matter of kinematics: you will determine if the cheetah’s high speed is enough to allow it to reach its prey before it runs out of steam (or time).
You will apply basic kinematics and simple assumptions to determine how a chase can play out.

Q2. In the elapsed time that the cheetah started and must stop, what distance can the gazelle cover?

 

(again clearly show your calculations)

 

 

Why is the 70 multiplied by 5/18?

Total time taken by cheetah = 13. 33 + 3.41
%3D
= 16.74 seconds
U sin g equation of motion for gazelle :
v2 – u² = 2x a' x d'
%3D
|
(70 x ) - 0 = 2×4.5 × d'
18
d'
= 42 m
%3D
u sin g equation of motion for gazelle :
v = u' + a't'
%3D
70 x = 0+ (4. 5 × r)
%3D
18
4.32 seconds
= t"
t'
Remaining time =
16. 74 4.32
= 12.42 seconds
u sin g equation of motion :
vxT = D
70 x x 12.42 = D
18
D = 241.5 m
Thus, total dis tan ce covered by gazelle in the time in which the cheetah comes
to a stop = 241.5 + 42 m
283.5 m
%3D
Transcribed Image Text:Total time taken by cheetah = 13. 33 + 3.41 %3D = 16.74 seconds U sin g equation of motion for gazelle : v2 – u² = 2x a' x d' %3D | (70 x ) - 0 = 2×4.5 × d' 18 d' = 42 m %3D u sin g equation of motion for gazelle : v = u' + a't' %3D 70 x = 0+ (4. 5 × r) %3D 18 4.32 seconds = t" t' Remaining time = 16. 74 4.32 = 12.42 seconds u sin g equation of motion : vxT = D 70 x x 12.42 = D 18 D = 241.5 m Thus, total dis tan ce covered by gazelle in the time in which the cheetah comes to a stop = 241.5 + 42 m 283.5 m %3D
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