SOLUTION: A lion is trying to catch an antelope. The antelope runs for 10.0 s at a constant velocity of +22.0 m/s. Starting from rest, what constant acceleration must the lion maintain in or
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Question 1172456: A lion is trying to catch an antelope. The antelope runs for 10.0 s at a constant velocity of +22.0 m/s. Starting from rest, what constant acceleration must the lion maintain in order to run the same distance as its prey(antelope) runs in the same time? (Assume that they both take a straight path) Found 2 solutions by math_tutor2020, Alan3354:Answer by math_tutor2020(3817) (Show Source):
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The antelope runs for 10.0 seconds at a constant velocity of 22.0 m/s
The distance the animal runs is 10*22 = 220 meters
The lion must run 220 meters in 10 seconds. It's starting velocity is vi = 0, since it's starting from rest. The lion's goal is to catch the antelope at the very tail end of the 220 meter stretch.
We can summarize what we know so far for the lion
x = distance = 220 meters
t = time = 10 seconds
vi = starting velocity = 0 m/s
We'll use this kinematics formula
x = vi*t + 0.5*a*t^2
Let's solve for 'a'
x = vi*t + 0.5*a*t^2
220 = 0*10 + 0.5*a*10^2
220 = 50a
50a = 220
a = 220/50
a = 4.4
The lion's acceleration must be 4.4 m/s^2, and this acceleration must be constant.
An acceleration of 4.4 m/s^2 means the lion's velocity is increasing by 4.4 m/s for each second.
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