Help With Position vs Time Graph

[theintarnets]

Homework Statement

I'm not sure if my graph for this problem is correct... I've attached it below, can someone please help me? The problem is:

In heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 minutes, then you have to stop for 3.5 minutes, and finally you drive at 15 m/s for another 2.5 minutes.

a) Plot the 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.

The Attempt at a Solution

Also, I'm not quite sure how to get average velocity out of this. I know that it's supposed to be distance divided by time, so I figured I could get the distance by multiplying 12 m/s by 90 seconds plus 15 m/s for 150 seconds to get 3330 meters, and then divide that by the total amount of time, 450 seconds, so my answer is 7.4 m/s but I don't know if that is right.

 

[oli4]

Hi the intarnets.
Now your graph is wrong, think about it,x can' go down because you never go backward.
The first part of the graph should start at x=0 and x will increase steadily because for each second passed, you will go "up" 12m, and this during 1.5 minutes
then x will stay the same for 3.5 minutes since you stopped, then... well, you get it now don't you ?

At the end of the graph, you will have a new x at its highest point
this will represent the total distance traveled, and you also know homw much time has passed, so you shouldn't have problem to find the average velocity (distance over time)

Cheers...

 

[azizlwl]

Assuming you stop the car in time approaches to zero, from 12 m/s to 0 m/s.

 

[ddnath]

All the information is not given to plot the graph and position vs time graph that you plotted is not correct.Here you tried to plot position in y-axis and time in x axis.But you notice,you plotted velocity in y-axis with respect to time.Moreover your information is not clear.You said,you drove for first 1.5 min (90 sec) with velocity 12m/s.During this time interval you traveled 90*12=1080m.here the you have to plot like this...(1,12),(2,24),(3,108)...up to (90,1080) by following the equation s=vt.Then you said you had to stop for 3.5 min.Its not clear that how long you needed to stop from 12m/s velocity to with what negative acceleration,how long you were stopped and how long you needed to gain 15m/s velocity from 0m/s with what acceleration.Unless you know these information,you can not draw the graph.Remember always that you are plotting position vs time graph,not velocity vs time graph.As you had not changed your direction so the value of y (position) may not increase for some time period (when you stopped),but will never decrease.

 

[theintarnets]

Thanks so much! My professor expects us to graph it with the information that's given. I tried the graph again, can someone check and see if it's correct? Also, I tried the average velocity again and I still got 7.4 m/s. Is that correct? Thanks again!

 

[HallsofIvy]

Yes, your calculation for the average speed is correct and your graph now looks good.