Hello M23 Originally Posted by

**M23** Here is the question...

**You are riding in an elevator. Starting from rest, the elevator undergoes the following motions:**

- It accelerates from rest (v=0.0 m/s) upwards for 5.0 at +2.0m/sē.
- It then coasts for 20.0 s at 10.0 m/s (a=0.0 m/sē).
- Finally, starting at 10.0 m/s, it accelerates downward for 2.5 s at -4.00 m/sē

Complete a position-time graph, a velocity-time graph, and an acceleration time graph.

**First, I completed a velocity-time graph, which looked something like this. **

But I don't exactly know how to make the position-time or the acceleration-time. Can someone explain how? Do I need a formula that finds distance using acceleration and time (for the position time graph)?

Any help would be appreciated. Thanks!

Your velocity-time graph is correct.

If you want a detailed distance-time graph, then you need to use the formula:

$\displaystyle s = ut +\tfrac12at^2$, giving the distance travelled, $\displaystyle s$, in terms of the initial speed, $\displaystyle u$, the acceleration $\displaystyle a$ and the time taken, $\displaystyle t$.

But if a sketch is all that's needed, then it will be in three sections:

- an parabola, starting horizontally and curving upward
- a straight line section - the tangent to the parabola at its final end-point
- a downward-curving parabola whose initial gradient is the same as that of the straight line and whose final gradient is zero (horizontal, when the lift has come to rest again)

The acceleration-time graph will be in three separate horizontal-line sections:

- one above the t-axis, representing the initial positive acceleration
- one along the t-axis: zero acceleration
- one below the t-axis: negative acceleration

Can you complete it now?

Grandad