Elevator Problem Distance Calculations
This picture was done with the help of 8th and 2nd Period Honors Physics at Manchester HS in 2018-19.
We're using the force calculations I did in the Elevator video to figure out a total distance the elevator runs from.
This is the first example of a multi-part problem, where a complex motion is best analyzed by breaking it into component pieces. The 4 pieces seen here are the same as they were on the chalkboard in the previous picture.
1.) Not moving
2.) Accelerating Downward
3.) Moving at constant speed
4.) Accelerating Upward (stopping)
Obviously, there's no motion during the first piece. During the next piece, the elevator accelerates downward and acquires a velocity v and a position x. Note that for the group whose data is on the board here they thought the acceleration began at 5 seconds and ended at 8 seconds, thus t=3.
The tricky part of this problem is recognizing that the v you get at the end of this downward acceleration is not only the final velocity for this part, but the original velocity v˳ of the next part. Similarly, the final position x of the downward acceleration is the original position x˳ of the next part. You carry the old x and v forward and make them the new x˳ and v˳
Elevator Problem Distance Calculations
This picture was done with the help of 8th and 2nd Period Honors Physics at Manchester HS in 2018-19.
We're using the force calculations I did in the Elevator video to figure out a total distance the elevator runs from.
This is the first example of a multi-part problem, where a complex motion is best analyzed by breaking it into component pieces. The 4 pieces seen here are the same as they were on the chalkboard in the previous picture.
1.) Not moving
2.) Accelerating Downward
3.) Moving at constant speed
4.) Accelerating Upward (stopping)
Obviously, there's no motion during the first piece. During the next piece, the elevator accelerates downward and acquires a velocity v and a position x. Note that for the group whose data is on the board here they thought the acceleration began at 5 seconds and ended at 8 seconds, thus t=3.
The tricky part of this problem is recognizing that the v you get at the end of this downward acceleration is not only the final velocity for this part, but the original velocity v˳ of the next part. Similarly, the final position x of the downward acceleration is the original position x˳ of the next part. You carry the old x and v forward and make them the new x˳ and v˳