An ant walks due east at a constant speed of 2 mi/hr on a sheet of paper that rests on a table. Suddenly, the sheet of paper starts moving due southeast at sqrt(2) mi/hr. The following is a correct solution for finding the resultant speed and direction of the ant relative to the table. Set up a coordinate system so that the positive y-axis corresponds to north and the positive x-axis corresponds to east. Then the velocity vector for the ant relative to the paper is given by <2, 0>. The velocity vector for the paper relative to the table is given by = <1, −1>. Therefore, the resultant velocity vector (which is the velocity vector of the ant relative to the table) is given by <2, 0> + <1, −1> = <3, −1>. The speed of the ant relative to the table is sqrt(3^2 + 1^2) = sqrt(10) mi/hr and the direction the ant is traveling relative to the table is found by using tan θ = −1/3, where θ represents the anglemade with the positive x-axis. In this case θ = arctan −1/3. In the following, you will solve two variations of this problem. In both variations, we will assume that the ant is walking on a paper that rests on a table, and then suddenly the sheet of paper starts moving. Make sure to model your work based upon what is provided here. Namely, make sure that you include a description of your coordinate system, that you represent any velocity vectors in terms of components, and that you relate any velocity vectors. (a) In the first variation, we want to find the velocity vector for the ant relative to the paper that would result in the ant traveling due east at 2 miles per hour relative to the table. The paper is still going to move due southeast at sqrt(2) mi/hr. (b) In the second variation, the paper is still going to move due southeast at sqrt(2) mi/hr. But now the ant is moving 2 mi/hr relative to the paper in a direction such that the ant travels due east relative to the table. Find the direction of travel of the ant relative to the paper. Also find the speed of the ant relative to the table.
An ant walks due east at a constant speed of 2 mi/hr on a sheet of paper that rests on a table. Suddenly, the sheet of paper starts moving due southeast at sqrt(2) mi/hr. The following is a correct solution for finding
the resultant speed and direction of the ant relative to the table.
Set up a coordinate system so that the positive y-axis corresponds to north and the positive x-axis corresponds to east. Then the velocity vector for the ant relative to the paper is given by <2, 0>. The velocity vector for the paper relative to the table is given by <sqrt(2) cos (-pi/4),sqrt(2 )sin−pi/4> = <1, −1>.
Therefore, the resultant velocity vector (which is the velocity vector of the ant relative to the table) is given
by <2, 0> + <1, −1> = <3, −1>. The speed of the ant relative to the table is sqrt(3^2 + 1^2) = sqrt(10) mi/hr and the direction the ant is traveling relative to the table is found by using tan θ = −1/3, where θ represents the anglemade with the positive x-axis. In this case θ = arctan −1/3.
In the following, you will solve two variations of this problem. In both variations, we will assume that the ant
is walking on a paper that rests on a table, and then suddenly the sheet of paper starts moving. Make sure to model your work based upon what is provided here. Namely, make sure that you include a description of
your coordinate system, that you represent any velocity vectors in terms of components, and that you relate
any velocity vectors.
(a) In the first variation, we want to find the velocity vector for the ant relative to the paper that would
result in the ant traveling due east at 2 miles per hour relative to the table. The paper is still going to
move due southeast at sqrt(2) mi/hr.
(b) In the second variation, the paper is still going to move due southeast at sqrt(2) mi/hr. But now the ant is moving 2 mi/hr relative to the paper in a direction such that the ant travels due east relative to the table. Find the direction of travel of the ant relative to the paper. Also find the speed of the ant relative to the table.
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