Average velocity between t= 0 and t = 7.5 minutes = 7.4 m / s
<h3>Further explanation </h3>Regular straight motion is the motion of objects on a straight track that has a fixed speed
Formula used
S = distance = m
v = speed = m / s
t = time = seconds
The motion of objects can be expressed in graphical form
This relationship graph can be in the form of a graph S-t, v-t or a-t
From the S-t graph, the average speed can be determined by determining the slope distance of the curve line from the distance to time ratio
average velocity = distance traveled / time taken
From the statement of questions there are 3 stages of time that occur
- 1. first time
move with v = 12 m / s, t = 1.5 min = 90 s
So the distance traveled =
s = v x t
S = 12. 90
S = 1080 m
- 2. second time
stop (v = 0) so that s = 0 and t = 3.5 min = 210 s
- 3. third time
move with v = 15 m / s and t = 2.5 min = 150 s
so the distance traveled
s = v x t
s = 15 x 150
s = 2250 m
Total distance = 2250 + 1080 = 3330 m
If we look at the existing chart (attached)
then we can draw a straight line from the starting point to the end point so we get a slash curve that shows the average speed
v average = distance traveled / time taken
v average = 3330/450
v average = 7.4 m / s
<h3>Learn more </h3>
The distance of the elevator
velocity position
resultant velocity
Keywords: average velocity, Motion of objects, s-t graph
