Answer:
A) 450 m
B) 27 m/s
C) 81 m, 243 m
D) Gazelle
Explanation:
A)
Since, the Cheetah is running at constant speed. Therefore, we use the equation:
s₁ = v₁t
where,
s₁ = distance covered by Cheetah = ?
v₁ = speed of Cheetah = 30 m/s
t = time taken = 15 s
Therefore,
s₁ = (30 m/s)(15 s)
<u>s₁ = 450 m</u>
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B)
For final speed of Gazelle at the end of 6 s acceleration time we use 1st equation of motion:
Vf = Vi + at
where,
Vf = Final Speed of Gazelle at the end of 6 s = ?
Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)
t = time taken = 6 s
a = acceleration = 4.5 m/s²
Therefore,
Vf = (0 m/s) + (4.5 m/s²)(6 s)
<u>Vf = 27 m/s</u>
C)
For the distance covered by Gazelle at the end of 6 s acceleration time we use 2nd equation of motion:
s₂ = Vi t + (0.5)at²
where,
Vi = Initial Speed of Gazelle = 0 m/s (Since, Gazelle is initially at rest)
t = time taken = 6 s
a = acceleration = 4.5 m/s²
Therefore,
s₂ = (0 m/s)(6 s) + (0.5)(4.5 m/s²)(6 s)²
<u>s₂ = 81 m</u>
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Now, for the distance covered during the last 9 s at constant velocity Vf, we use equation:
s₃ = Vf t
where,
s₃ = distance covered by Gazelle in last 9 s = ?
t = time = 9 s
Therefore:
s₃ = (27 m/s)(9 s)
<u>s₃ = 243 m</u>
D)
We know that, at the end of 15 seconds:
Distance covered by Cheetah = s₁ = 450 m
Distance Covered by Gazelle = s₂ + s₃ = 81 m + 243 m = 324 m
If we take the initial position of Cheetah as origin. Then the positions of both Gazelle and Cheetah with respect to origin will be:
Position of Cheetah = 450 m ahead of origin
Position of Gazelle = 324 m + 160 m = 484 m (Since, Gazelle was initially 160 m ahead of Cheetah)
<u>Hence, it is clear that Gazelle is ahead at the end of 15 s.</u>
