Question

13.) write everything you know from the distance vs. time graph below: (minimum of 5, 6 - 10 are extra credit!) part description 1.) ______ ____ 2.) ____ ____ 3.) ____ ____ 4.) ____ ____ 5.) ____ ____ 6.) ____ ____ 7.) ____ ____ 8.) ____ ____ 9.) ____ ____ 10.) ____ ______

Explanation:

Step1: Analyze Part A

The object starts at a distance of 2 units from the origin at time t = 0. It moves with a constant - positive velocity as the distance - time graph has a positive slope. The slope of the line in Part A gives the speed of the object during this time interval.

Step2: Analyze Part B

The object reaches a maximum distance of 4 units at time t = 2. During this part, the object is still moving in the positive direction but the speed is decreasing as the slope of the graph is decreasing.

Step3: Analyze Part C

From t = 2 to t = 4, the object is at rest. The distance does not change with time, so the slope of the graph is zero.

Step4: Analyze Part D

From t = 4 to t = 8, the object moves back towards the origin. The slope of the graph is negative, indicating a negative velocity. The object is moving with a constant speed during this part as the slope of the line is constant.

Step5: Calculate speed in Part A

The initial point of Part A is (0,2) and the end - point is (2,4). The slope $m=\frac{\Delta x}{\Delta t}=\frac{4 - 2}{2-0}=1$ unit per time unit. So the speed in Part A is 1 unit per time unit.

Step6: Calculate speed in Part D

The initial point of Part D is (4,2) and the end - point is (8,0). The slope $m=\frac{\Delta x}{\Delta t}=\frac{0 - 2}{8 - 4}=-\frac{1}{2}$ unit per time unit. The speed is $\frac{1}{2}$ unit per time unit (speed is non - negative, we take the magnitude of the slope).

Step7: Total displacement

The object starts at x = 2 and ends at x = 0. So the total displacement is $0 - 2=-2$ units.

Step8: Total distance traveled

In Part A, the object moves $4 - 2 = 2$ units. In Part B, it moves $4-4 = 0$ units (just changes its rate of motion). In Part C, it moves 0 units. In Part D, it moves $2-0 = 2$ units. So the total distance traveled is $2+0 + 0+2=4$ units.

Step9: Average velocity

Average velocity $v_{avg}=\frac{\text{Displacement}}{\text{Total time}}=\frac{-2}{8}=-\frac{1}{4}$ unit per time unit.

Step10: Average speed

Average speed $s_{avg}=\frac{\text{Total distance}}{\text{Total time}}=\frac{4}{8}=\frac{1}{2}$ unit per time unit.

Answer:

1. The object starts at a distance of 2 units from the origin at t = 0.
2. In Part A (0 < t < 2), the object moves with a constant positive velocity of 1 unit per time unit.
3. The object reaches a maximum distance of 4 units at t = 2.
4. In Part B (2 < t < 2), the object is still moving in the positive direction but with decreasing speed.
5. In Part C (2 < t < 4), the object is at rest.
6. In Part D (4 < t < 8), the object moves back towards the origin with a constant speed of $\frac{1}{2}$ unit per time unit.
7. The total displacement of the object is - 2 units.
8. The total distance traveled by the object is 4 units.
9. The average velocity of the object is $-\frac{1}{4}$ unit per time unit.
10. The average speed of the object is $\frac{1}{2}$ unit per time unit.