+16 Starting from rest at the ground floor, an elevator takes 19.0 seconds for a vertical trip of 70.0 meters. It begins its journey with 4.50 seconds of constant acceleration, then moves for 9.00 seconds at constant velocity, and finally, moves for 5.50 seconds of constant negative acceleration that brings the elevator to a stop. (a) What is the maximum velocity of the elevator in m/s?
+118613 Hi Jasmine, I like this question - I had to think a bit. :)
Starting from rest at the ground floor, an elevator takes 19.0 seconds for a vertical trip of 70.0 meters. It begins its journey with 4.50 seconds of constant acceleration, then moves for 9.00 seconds at constant velocity, and finally, moves for 5.50 seconds of constant negative acceleration that brings the elevator to a stop. (a) What is the maximum velocity of the elevator in m/s?
t=19
s=70
4.5sec then 9 sec then 5.5sec
Here are the satandard motion in a straight line equations
I only used equation 3 And I let v be the maximum velocity
Acceleration section
\(s_1=0.5(0+v)*4.5\\ s_1=0.5*4.5v\\ s_1=2.25v\)
middle section
\(s_2=0.5(v+v)*9\\ s_2=9v\\\)
deceleration section
\(s_3=0.5(v+0)*5.5\\ s_3=0.5*5.5v\\ s_3=2.75v\\\)
Now the total distance traveled is 70m
\(s_1+s_2+s_3=70\\ 2.25v+9v+2.75v=70\\ 14v=70\\ v=5m/s\)
Maximum velocity = 5m/s
