Question:

The position-time graph shown below represents the motion of an object over a period of 10 seconds. Use the graph to answer the following questions. (Assume positive position means movement in the positive direction.)

a) Determine the velocity of the object between t = 2 s and t = 6 s.

b) Calculate the average acceleration of the object between t = 4 s and t = 8 s.

c) Identify the time interval(s) where the object is at rest.

d) Determine the total displacement of the object over the entire 10-second period.

Answer:

a) The velocity of an object corresponds to the slope of the position-time graph. To determine the velocity between t = 2 s and t = 6 s, we'll calculate the slope of the line segment connecting these two points.

Using the slope formula: [ \text{slope} = \frac{{\text{change in y-coordinate}}}{{\text{change in x-coordinate}}} ]

We find: [ \text{slope} = \frac{{y_2 - y_1}}{{x_2 - x_1}} = \frac{{60 , \text{m} - 20 , \text{m}}}{{6 , \text{s} - 2 , \text{s}}} = \frac{{40 , \text{m}}}{{4 , \text{s}}} = 10 , \text{m/s} ]

Therefore, the velocity of the object between t = 2 s and t = 6 s is 10 m/s.

b) Average acceleration corresponds to the rate of change of velocity. To calculate the average acceleration between t = 4 s and t = 8 s, we'll determine the change in velocity and divide it by the corresponding time interval.

Between t = 4 s and t = 8 s, the change in velocity is equal to the change in slope of the position-time graph. We'll calculate the slope at both t = 4 s and t = 8 s.

At t = 4 s: [ \text{slope}1 = \frac{{y{2,1} - y_{1,1}}}{{x_{2,1} - x_{1,1}}} = \frac{{0 , \text{m/s} - 20 , \text{m/s}}}{{8 , \text{s} - 4 , \text{s}}} = \frac{{-20 , \text{m/s}}}{{4 , \text{s}}} = -5 , \text{m/s}^2 ]

At t = 8 s: [ \text{slope}2 = \frac{{y{2,2} - y_{1,2}}}{{x_{2,2} - x_{1,2}}} = \frac{{40 , \text{m/s} - 60 , \text{m/s}}}{{8 , \text{s} - 4 , \text{s}}} = \frac{{-20 , \text{m/s}}}{{4 , \text{s}}} = -5 , \text{m/s}^2 ]

The average acceleration between t = 4 s and t = 8 s is the average of these two slopes: [ \text{average acceleration} = \frac{{\text{slope}_1 + \text{slope}_2}}{2} = \frac{{-5 , \text{m/s}^2 + (-5 , \text{m/s}^2)}}{2} = -5 , \text{m/s}^2 ]

Therefore, the average acceleration of the object between t = 4 s and t = 8 s is -5 m/s^2.

c) The object is at rest when its velocity is zero. From the graph, we can observe that the object is at rest between t = 0 s and t = 2 s, and between t = 6 s and t = 10 s.

Therefore, the time intervals where the object is at rest are t = 0 s to t = 2 s, and t = 6 s to t = 10 s.

d) The total displacement of an object can be determined by finding the area under the velocity-time graph. In this case, since we have a position-time graph, we can find the area under the graph between t = 0 s and t = 10 s.

The area under a position-time graph represents displacement, which is equal to the rectangle's width times its height. The width of the rectangle is 10 s, and the height is the average velocity:

Now, we can calculate the total displacement:

Therefore, the total displacement of the object over the entire 10-second period is 60 meters.

(Note: The negative area under the graph represents any motion in the negative direction, but since the graph doesn't cross the time axis, there is no negative displacement in this case).