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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1) X-component: 568.5 N, Y-component: 511.9 N
2) The horizontal force is 86.6 N and the resulting acceleration is
Explanation:
1)
In this first part of the problem, we have to resolve the force into its two components, along the x and the y direction.
The two components are given by:
where
F = 765 N is the magnitude of the force
is the angle of the force with the horizontal
Substituting, we find:
- Horizontal component:
- Vertical component:
2)
First of all, we have to find the horizontal component of the pulling force, which is given by
where
F = 100 N is the magnitude of the pulling force
is the direction of the force with the horizontal
Substituting,
Now we can find the acceleration of the wagon by using Newton's second law:
where
is the net force in the horizontal direction
m = 100 kg is the mass of the wagon
is the acceleration in the horizontal direction
Solving for , we find
Learn more about vector components:
And about Newton's second law:
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