In heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 milrutes, then you have to stop for 3.5 minutes, and finally i you drive at 15 m/s for another 2.5 minutes. (a) plot a position-versus-time graph for this motion. your plot should extend from t= 0 to t =7 .5 minutes. (b) use your plot from part (a) to calculate the average velocity between t= 0 and t = 7.5 minutes.

Average velocity  between t= 0 and t = 7.5 minutes = 7.4 m / s

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

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

stop (v = 0) so that s = 0 and t = 3.5 min = 210 s

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

The distance of the elevator

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velocity position

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resultant velocity

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Keywords: average velocity, Motion of objects, s-t graph

prolly a beacuse the momentuem the car has increases while going down hill