A person on horseback moves according to the velocity-versus-time graph shown in the figure (attached) . . A. Find the displacem
ent of the person for segment A of the motion. . . B. Find the displacement of the person for segment B of the motion. . . C. Find the displacement of the person for segment C of the motion.
1 answer:
From my research, the following image is used for the problem. For each segment, the displacement can be obtained by looking for the area under the segment.
Part A:
displacement = (1/2)*10*2 = 10m
Part B:
displacement = [(1/2)*4*5] + (5*2) = 20m
<span>Part C:
</span>
displacement = [(1/2)*4*10] + (2*10) = 40m
Part A:
displacement = (1/2)*10*2 = 10m
Part B:
displacement = [(1/2)*4*5] + (5*2) = 20m
<span>Part C:
</span>
displacement = [(1/2)*4*10] + (2*10) = 40m
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Answer:
(a) 0 J
(b) 828.96 J
(c) 828.96 J
(d) 828.96 J
(e) 0 J
(f) 828.96 J
Explanation:
Given:
Mass of the rock is,
Initial height of the rock is,
Final height of the rock is,
Initial velocity of the rock is,
Acceleration due to gravity is,
(a)
Initial kinetic energy is given as:
Plug in the given values and solve for ''. This gives,
Therefore, initial kinetic energy is 0 J.
(b)
Initial potential energy is given as:
Therefore, initial potential energy is 828.96 J.
(c)
Mechanical energy is equal to the sum of kinetic and potential energy. It is always a constant value. Therefore,
Therefore, mechanical energy at 26.6 m is 828.96 J.
(d)
Now, at the final height of 0 m, the decrease in potential energy will be equal to the increase in the kinetic energy. In other words, the potential energy at the start will be converted to kinetic energy at the bottom.
Therefore, kinetic energy at the 0 m is given as:
(e)
Potential energy at the final height is given as:
Therefore, final potential energy is 0 J.
(f)
Total mechanical energy is a constant and doesn't depend on the height.
So, total mechanical energy at 0 m is same as that at 26.6 m.
Total mechanical energy is 828.96 J.