uniform motion graph

Question

A squirrel has just grabbed a peanut from an unsuspecting park visitor. In his panicked attempt to get safely back to his tree, he runs back and forth a few times to elude any would be pursuers. A graph of his motion is shown below.

link to graph https://media.cheggcdn.com/media/ca6/ca68f398-6684-442d-bfce-8b62e5bfd69c/phptXDlM8.png

1. During which time interval is the squirrel not moving?

2. During which time interval is the squirrel moving forwards the fastest?

3. During which time interval is the squirrel moving backwards the slowest?

4. How far away is the tree from where the squirrel grabbed the peanut?

5. What is the squirrel’s average velocity on his run to the tree?

6. What distance does the squirrel actually run?

7. What is the squirrel’s average speed for his run to the tree?

Answer 1

Motion/Velocity analysis:

Forwards motion (velocity) on time intervals A, D, and G

Backwards motion (velocity) on time intervals B, E, and F

No motion (zero velocity) on time interval C

1. C

2. G (steepest positive slope)

3. F (steepest negative slope)

4. 11 meters (at t = 0, the position = 0. This is where the peanut was grabbed. At t = 20, the position is at 11 meters. This is the tree. 11 meters from the grab point.

5. Average velocity = (11 m)/(20 s) = 0.55 m/s

6. Total distance = A + B + C + D + E + F + G = 6 + 2 + 0 + 6 + 3 + 4 + 8 = 29 meters

7. Average speed = (29 m)/(20 s) = 1.45 m/s