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MMH The earth moves around the sun in a nearly circular orbit of radius $1.50 \times 10^{11} {m}$ During the three summer months (an elapsed time of $7.89 \times 10^{6} {s} )$ the earth moves one-fourth of the distance around the sun. (a) What is the average speed of the earth? (b) What is the magnitude of the average velocity of the earth during this period?
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Physics 101 Mechanics
Chapter 3
Kinematics in Two Dimensions
Motion in 2d or 3d
Rutgers, The State University of New Jersey
Simon Fraser University
Hope College
McMaster University
Lectures
04:01
2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.
10:12
A vector is a mathematical entity that has a magnitude (or length) and direction. The vector is represented by a line segment with a definite beginning, direction, and magnitude. Vectors are added by adding their respective components, and multiplied by a scalar (or a number) to scale the vector.
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The earth moves around the…
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The average distance betwe…
04:10
Interactive Solution $\und…
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(a) Calculate Earth's…
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Interactive Solution $3.1…
06:20
The radius of the earth…
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Every year the Earth trave…
02:06
The earth orbits the sun o…
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So the question states, we need to find the speed and the velocity of the earth as it travels 1/4 of its orbit. So to find the speed we need to find what the distance is that it travels along its orbit. And to do this we know that the angle here is 90 degrees because it travels 1/4 of its orbit and we also know the radius was Call it are for now. But it's stated up here to find this distance we just needed, multiply are and high over to Now that we know this distance, we can divide by time t. And when we do this, we get an answer of 29,000 863 meters per second. Now to find the velocity, all we need to find is the displacement between the initial and the final point. So here's the initial point. Here's the final point, and we need to find this distance here. We can do this because it makes a triangle with both sides being of radius R. So that means the distance is going to be equal to square root of R squared plus R squared and then we can divide by the time to give us the velocity and when we do this, it's the same thing. A square root of two are divided by t, and when we plug everything in that we know we get 28,000 886 meters per second.
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