Thread: Graphing!

A cart moves to the right of the origin at 3.6m/s for 8s. It stops abruptly and remains stationary for 4 seconds. The cart then moves back towards the origin at 1.8m/s. Choose motion to the right to be in the positive direction.

I really do not under stand what "Choose motion to the right to be in the positive direction" and would this graph start at (0s,0m)? Can some one please give me a sketch of how it would look like? Thank you!

I assume you were asked to plot velocity vs time, right?

If that's the case I guess you could draw 3 horizontal lines. The first at v=3.6 m/s from 0 to 8s. The second at v=0 from 8 to 12s. And the third one at v=-1.8 m/s starting at t=12s.

Here's a graph of your model.



I just noticed the first point plotted reads 8*3.8, but should be 8*3.6. Here's the function that was plotted:

The third one with v= -1.8m/s how would i plot it? would i assume 12s = 0s, 13s=1 s or how?

is it's equation of the line y=-1.8x ? because it also supposed to go to the origin?

and how did you calculate it would take 28s for the cart to return?

and what would be the relation ship between the magnitude of the slope and of the velocity?

this is my first time doing this >.<!

Hello, sakonpure6!

The instructions are incomplete.

A cart moves to the right of the origin at 3.6m/s for 8s.
It stops abruptly and remains stationary for 4 seconds.
The cart then moves back towards the origin at 1.8m/s.
Choose motion to the right to be in the positive direction. . Then what?

We can only guess what the problem wants us to do.


[1] We can construct a piecewise function of the cart's position at time

. . .



[2] We can sketch a graph of the cart's position at time

Code:

      |
   288+       * * *
      |      *      *
      |     *         *
      |    *            *
   144+   *               *
      |  *                  *
      | *                     *
      |*                        *
    --*---+---+---+---+---+---+---*----
      |   4   8  12  16  20  24  28
      |