00:01
We have to solve the following problem, which describes the journey of the worker from the ground floor to the floor where she works.
00:11
So on the first, let's look at the sketch for the acceleration.
00:16
For the first five seconds, acceleration is zero.
00:27
And we presume that this time is used for the worker to come into the elevator.
00:36
Four elevator is not moving as well and the coordinate is also zero so let's sketch let's sketch this stage here so i'll try to do my best in sketching so uh here both height and velocity are zero then from five to ten seconds it accelerates with a constant accelerate of 4 meters per second square.
01:23
During this time the velocity of the elevator increases from zero to final to some final value and this final value equals to a t where time is a total time of acceleration.
01:46
So the initial velocity is zero.
01:49
The final velocity equals to acceleration times 10 seconds minus 5 seconds or 4 meters per second squared times 5 seconds which is 20 meters per second now let's catch it here so here for the simplicity let's say that one step is 5 meters 5 meters per second square then this is 10 15 20s here so this is a line i hope it looks like a line now let's write down now let's determine an equation oh sorry let's determine the equation for the displacement during first five seconds the displacement or height of the elevator equals to h0 plus v0 t plus a t squared over 2.
03:22
And the first two terms equals to 0.
03:27
Therefore it equals to 80 squared over 2 or 2 times t squared meters.
03:39
So let's calculate h as a function of 10 minus 5.
03:54
So here obviously time is shifted.
03:57
So this should be labeled as t prime, which equals to, therefore it equals to a or 2 times t minus 5 squared.
04:16
So let's show it...