# A position-time graph and an acceleration-time graph.

• Feb 15th 2010, 09:36 PM
M23
A position-time graph and an acceleration-time graph.
You are riding in an elevator. Starting from rest, the elevator undergoes the following motions.

It accelerates from rest (v=0.0m/s) upwards for 5.0 s at +2.0m/s²

It then coasts for 20.0s at 10.0m/s (a=0.0m/s²)

Finally, starting at 10.0m/s, it accelerates downward for 2.5s at -4.00m/s²

I did the velocity time graph just fine. I am, however, at a complete loss on how to do the position time and acceleration-time ones.
Are there any calculations I have to do?

Any help would be much appreciated.
• Feb 16th 2010, 06:51 AM
mathemagister
Quote:

Originally Posted by M23
You are riding in an elevator. Starting from rest, the elevator undergoes the following motions.

It accelerates from rest (v=0.0m/s) upwards for 5.0 s at +2.0m/s²

It then coasts for 20.0s at 10.0m/s (a=0.0m/s²)

Finally, starting at 10.0m/s, it accelerates downward for 2.5s at -4.00m/s²

I did the velocity time graph just fine. I am, however, at a complete loss on how to do the position time and acceleration-time ones.
Are there any calculations I have to do?

Any help would be much appreciated.

Here's a kick start:
For the acceleration-time graph, just see what acceleration the elevator is going at. So, for the first 5 seconds plot the acceleration as a horizontal line at +2.0m/s². Then, for 20 seconds, the line jumps down to the axis (0.0m/s²). Then, for 2.5s, plot the acceleration as a horizontal line at -4.00m/s². So, the acceleration-time graph should just be 3 horizontal lines.
• Feb 16th 2010, 07:02 AM
Hello M23
Quote:

Originally Posted by M23
You are riding in an elevator. Starting from rest, the elevator undergoes the following motions.

It accelerates from rest (v=0.0m/s) upwards for 5.0 s at +2.0m/s²

It then coasts for 20.0s at 10.0m/s (a=0.0m/s²)

Finally, starting at 10.0m/s, it accelerates downward for 2.5s at -4.00m/s²

I did the velocity time graph just fine. I am, however, at a complete loss on how to do the position time and acceleration-time ones.
Are there any calculations I have to do?

Any help would be much appreciated.

On the position-time graph, the gradient (slope) of the graph represent the velocity at that instant. So, since the initial velocity is zero, the position-time graph starts with zero gradient - in other words, it's horizontal.

Then, during the accelerating phase between t = 0 and t = 5, the gradient of the position-time graph increases until the maximum velocity is reached. So this section of the graph will be an increasingly steep curve which has the t-axis as a tangent at the origin.

When the lift is moving with a constant velocity, between t = 5 and t = 25, the displacement-time graph continues to move upwards in a diagonal straight line.

As the lift decelerates to rest, the graph is once more a curve, but this time with a decreasing gradient, finally becoming horizontal when the lift stops moving.

The overall shape of the position-time graph, then, is rather like a letter S, which has been stretched out horizontally.

The acceleration time graph just consists of three horizontal-line sections: the first, between t = 0 and t = 5, the horizontal line where a = 2; then between t = 5 and t = 25, a section of the t-axis itself (that's where a = 0); finally between t = 25 and t = 27.5 a horizontal line below the t-axis, where a = -4.

Can you draw these graphs now?

• Feb 16th 2010, 02:32 PM
M23
Thank you both very much. I think I've got both graphs done, now. The part with the "horizontally stretched S" was particularly helpful, thank you.

Oh, just a quick question I thought I might add, for the accel-time graph, would I draw vertical lines between the horizontal ones?
• Feb 16th 2010, 02:45 PM
mathemagister
Quote:

Originally Posted by M23
Thank you both very much. I think I've got both graphs done, now. The part with the "horizontally stretched S" was particularly helpful, thank you.

Oh, just a quick question I thought I might add, for the accel-time graph, would I draw vertical lines between the horizontal ones?

No, that is unconventional, because that would mean that, at that time, you were accelerating at an infinite number of different accelerations simultaneously, so just leave them disconnected. To make it a real function, it can't have 2 values at the changing points, so it would make sense to draw a blank hole on the previous horizontal line's end, and draw a full hole at the start of the next one. Do you get what I said?

So it would look something like (ignore the *s):
-----------------O

************** $\cdot$---------------------