Help finding slope on position-time graph

[ultra_mc]

Im trying to make fill in a velocity-time graph from the a picture of a position-time graph. The position-time graph starts at 20m (no direction indicated) and slides dowb to 5m in 4 seconds. The slope would be -4m/s right? But the V-T graph I have to fill in has a maximum velocity of 6 m/s and a min of -6 m/s. What is the corect way of finding this?

PS. while typing this I might have found out. Would you just draw a straight line starting at 4m/s until 4 seconds?

Edit: Then I would have to convert the V-T graph to an A-T graph, using the slope of the V-T graph (which I think would be 0) how would that be drawn on the A-T graph?

^^^That would be a straight line at 0 right?

 

[jbsteele]

Yes, your solution is correct. The V-T graph would consist of a straight line starting at 4 m/s and going to 0 m/s after 4 seconds. To convert the V-T graph to an A-T graph, you would draw a straight line at 0, as you mentioned. This would indicate that the acceleration is 0.

 

[tomcue]

Dear student,

Thank you for reaching out for help with finding the slope on a position-time graph. It seems like you have a good understanding of the concept, but let me clarify a few things for you.

Firstly, you are correct in saying that the slope of the position-time graph would be -4m/s. This is because the slope represents the rate of change of position over time, and in this case, the position is changing by 15m (20m to 5m) in a time of 4 seconds. So the slope would be -15m/4s which simplifies to -4m/s.

Now, when it comes to filling in the velocity-time graph, you are correct that you would start at a velocity of 4m/s (since the slope is -4m/s) and draw a straight line until 4 seconds. However, to make the graph more accurate, you would need to consider the fact that the velocity is changing at a constant rate. This means that the velocity would decrease by 1m/s every second, resulting in a final velocity of 0m/s at 4 seconds. So the graph would look like a straight line decreasing from 4m/s to 0m/s.

Now, to answer your question about converting the velocity-time graph to an acceleration-time graph, you are correct in saying that the slope of the velocity-time graph would be 0m/s^2. This is because the velocity is not changing over time, so the rate of change (i.e. the slope) would be 0. Therefore, the acceleration-time graph would be a straight line at 0m/s^2.

I hope this helps clarify the process for you. Remember, when working with graphs and finding slopes, it is important to always pay attention to the units and the direction of the graph. Keep up the good work!