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A cheetah is hunting. Its prey runs for 3.0 $\mathrm{s}$ at a constant velocity of $+9.0 \mathrm{m} / \mathrm{s}$ . Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs inthe same time?
$a=6 \mathrm{m} \cdot \mathrm{s}^{-2}$
Physics 101 Mechanics
Chapter 2
Kinematics in One Dimension
Motion Along a Straight Line
University of Washington
Simon Fraser University
McMaster University
Lectures
04:34
In physics, kinematics is …
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the cheetah is hunting some prey and the prey, and this problem has a speed or velocity of nine meters per second and travels for three seconds. So the pray, whatever it is, travels. We used the equation velocity times, time that prey travels nine times three or 27 meters. So the question asks, How fast would a cheetah have to accelerate if it starts from rest? Be not Cheetah zero and travels the same distance. So that would also be 27 in the same amount of time. So we can use the expression X equals B, not t plus 1/2 A t squared because our cheetah is accelerating. So 27 meters equals initial velocity, which is zero times three plus 1/2 a times three Where so simplifying this a little bit we don't that first term goes away. We just have 27 meters equals 1/2 a t squared so we can say 54 meters is equal to a times three squared three squared is nine. So dividing 54 by nine our acceleration is six meters per second squared
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