Question 27 - Problems - Chapter 2¶
Problem:¶
An elevator travels up from the first floor to the ninth floor in 20 s, down from the ninth floor to the fifth floor in 12 s, and finally up from the fifth floor to the twelfth floor in 18 s.
The spacing between any two adjacent floors is 4.0 m.
Question:¶
For the entire period of 50 s, what is the average speed? What is the average velocity?
Solution:¶
$\bigtriangleup t_1 = 20s$
$\bigtriangleup t_2= 12s$
$\bigtriangleup t_3= 18s$
One floor = 4m
$\bigtriangleup s_1 = 8x4m=32m$
$\bigtriangleup s_2 = 4x4m=-16m$
$\bigtriangleup s_2 = 7x4m=28m$
The total displacement $\bigtriangleup s = \bigtriangleup s_1 + \bigtriangleup s_4 + \bigtriangleup s_5$
The average velocity:
$$
\overline v=\frac{\bigtriangleup s}{\bigtriangleup t}
$$
Total distance $s = \Big|{{\bigtriangleup s_1}}\Big| + \Big|{\bigtriangleup s_2}\Big| + \Big|{\bigtriangleup s_3}\Big|$
The average speed is
$$
v = \frac{s}{t}
$$