placement, Velocity, and Acceleration Work out the following problems. Be sure to show your work in detail. See the uploaded files for examples of how the problems should be solved and presented. A car is driving along a straight section of highway between Oklahoma City and Tulsa. At one point in time, the car is 50 miles from Tulsa. At a later point in time, it's 40 miles from Tulsa. What is its displacement? s = x-x0 s= 50 40 s=10 miles 2. An aircraft flies 20 km on a course of 030 degrees, then 30 km on a course of 270 degrees. What is its distance and bearing from its takeoff point? (Hints: Make a sketch. Convert flight legs to Cartesian vectors in standard position. See the "Overview of Vectors" handout.) 20 km at 30O 30 km at 270O VR (XRYR)= V1 = V2 Because the second vector is at 270O it makes a right angle with the first vector, so that the length of the resulting vector, VR can be determined with the simple equation a2 + b2= c2 202 + 302 = c2 400+900= c2 c= 36.06km Then, one can determine the angle of a, the angle including the 30O of the initial angle and whatever the angle is below the x-axis by looking at the sine, which is the opposite leg (30km) over the hypotenuse (36km). sin O of a= 30/ 36.06 = 8.32, therefore a= 56.30 O 56.30 30 = 26.30 O Therefore, the vector is traveling at 26.30 O clockwise from the x-axis, but coordinates are measured counter-clockwise in the Cartesian plane, so 360 O 26.30 O = 333.77 O Therefore, the
Once you have finished the video let’s practice by completing these problems (be sure to include units!):
Draw a graph that shows the distance Jacob’s car is from his house with respect to time. Remember
For problems 12 to 14, do the following: (a) Make a scatter diagram of the
reduce the sides and then use my first equation and It worked! So I tried it
Answer the questions below. When you are finished, submit this assignment to your teacher by the due date for full credit.
Answer each problem thoroughly. Each problem is worth 10 points. Make sure to provide detailed explanations for each part of each problem.
a.Calculate the velocity of the ship on the river. vwater = (-5, 0)m/s, v = v0 + vwater =Answer (7, 6) m/s
41328.125 a = 41328.125 m/s^2 Then we take v = u +at where v is the final velocity of an object, u is the initial velocity, a is the acceleration and t is the time from u to v. u=0 v=230 m/s a=41328.125 m/s^2 230= 0 + 41328.125 x
The resultant would go between so the plane will not be heading the west direction, therefore the airplane should fly at an angle to counter the wind to arrive at it destination which is due west.
To use this shortcut, you need to measure the 100 yard side, which costs $2,500, the 60° angle, which costs $4,500, and the 200 yard side, which costs $5,000. The total cost of this shortcut is $12,000, which is exactly at the budget. The second shortcut also uses SAS theorem. To use this shortcut you need to measure the 173 yard side, which costs $4,325, the 30° angle, which costs $2,250, and the 200 yard side, which costs $5,000. That means this shortcut’s total cost is $11,575. The next shortcut uses ASA theorem, and it requires you to measure the 30° angle, which costs $2,250, the 200 yard side, which costs $5,000, and the 60° angle, which costs $4,500, for a total cost of $11,750. Another within budget shortcut uses the AAS theorem. This shortcut requires you to find the measures of the 30° angle, the 90° angle, and the 100 yard side, which costs $2,500, $6,750, and $2,500 respectively. That means this shortcut has a total cost of $11,500. The next shortcut also uses AAS. To use it you need to measure the 30° angle, which costs $2,250, the 60° angle, which costs $4,500, 173 yard side, which costs $4,325. In total this shortcut costs $11,075. Another shortcut uses SSS theorem, and it requires for you to measure every side. The 173 yard side costs $4,325, the 100 yard side costs $2,250, and the 200 yard side costs $5,000, making the total cost
For this situation the measured stopping distance is 302.28 feet and the velocity at which the brakes are applied is 88
The derivative of the flight path angle must be negative, it has to decrease. For that reason, if dγ/dt is positive, the theta angle will be computed in order to impose a null derivative. Then, this theta angle is taken into account in all the derivative equations.
From looking at this formula it leads me to believe to calculate y = (x + b), and not y = (mx + b). However,
6. A positive angle of attack up to and beyond the wing design limit will completely destroy
Where the is the azimuth angle measure from the north clockwise, is the zenith distance that mean the angle the vertical and the radian and r is the radian of the in the local system.