7. given the following verbal description draw a position vs. time graph. an object starts at 5 meters and…

7. given the following verbal description draw a position vs. time graph. an object starts at 5 meters and moves to the left at constant speed for 3 seconds. the object stops for 2 seconds and then continues in the same direction and speed as before for 2 more seconds. the object stops for 2 seconds and then moves to the right for 2 seconds at a greater speed. 8. from the position vs. time data below, answer the following questions. a. construct a graph of position vs. time. b. construct a graph of velocity vs. time. t (s) x (m) 0 0 1 2 2 4 3 4 4 7 5 10 6 10 7 10 8 5 9 0 c. determine the displacement from t = 3.0s to 5.0s using graph b. check it by using graph a. d. determine the displacement from t = 7.0 s to 9.0 s using graph b. check it using graph a

7. given the following verbal description draw a position vs. time graph. an object starts at 5 meters and moves to the left at constant speed for 3 seconds. the object stops for 2 seconds and then continues in the same direction and speed as before for 2 more seconds. the object stops for 2 seconds and then moves to the right for 2 seconds at a greater speed. 8. from the position vs. time data below, answer the following questions. a. construct a graph of position vs. time. b. construct a graph of velocity vs. time. t (s) x (m) 0 0 1 2 2 4 3 4 4 7 5 10 6 10 7 10 8 5 9 0 c. determine the displacement from t = 3.0s to 5.0s using graph b. check it by using graph a. d. determine the displacement from t = 7.0 s to 9.0 s using graph b. check it using graph a

Answer

Answer:

7.

To draw the position - vs - time graph:

  1. Start the object at (x = 5) m at (t=0). For the first 3 seconds, since it moves to the left at a constant speed, the position - time graph is a straight - line with a negative slope.
  2. Then it stops for 2 seconds, so the graph is a horizontal line from (t = 3) s to (t=5) s.
  3. It continues in the same direction at the same speed for 2 more seconds, so it's another straight - line with the same negative slope as before from (t = 5) s to (t = 7) s.
  4. Then it stops for 2 seconds, so a horizontal line from (t = 7) s to (t = 9) s.
  5. Finally, it moves to the right for 2 seconds at a greater speed, so a straight - line with a positive and steeper slope from (t = 9) s to (t = 11) s.

8.

a.

The position - vs - time graph is already partially drawn in the given figure. Plot the points from the table ((t,x)): ((0,0)), ((1,2)), ((2,4)), ((3,4)), ((4,7)), ((5,10)), ((6,10)), ((7,10)), ((8,5)), ((9,0)) and connect them with straight - line segments.

b.

To construct the velocity - vs - time graph:

  1. For (t = 0) to (t = 1) s, (v=\frac{\Delta x}{\Delta t}=\frac{2 - 0}{1-0}=2) m/s.
  2. For (t = 1) s to (t = 2) s, (v=\frac{4 - 2}{2 - 1}=2) m/s.
  3. For (t = 2) s to (t = 3) s, (v=\frac{4 - 4}{3 - 2}=0) m/s.
  4. For (t = 3) s to (t = 4) s, (v=\frac{7 - 4}{4 - 3}=3) m/s.
  5. For (t = 4) s to (t = 5) s, (v=\frac{10 - 7}{5 - 4}=3) m/s.
  6. For (t = 5) s to (t = 6) s, (v=\frac{10 - 10}{6 - 5}=0) m/s.
  7. For (t = 6) s to (t = 7) s, (v=\frac{10 - 10}{7 - 6}=0) m/s.
  8. For (t = 7) s to (t = 8) s, (v=\frac{5 - 10}{8 - 7}=- 5) m/s.
  9. For (t = 8) s to (t = 9) s, (v=\frac{0 - 5}{9 - 8}=-5) m/s. Plot these velocity values at the mid - points of the time intervals.

c.

From graph A (position - vs - time graph): At (t = 3.0) s, (x_1 = 4) m. At (t = 5.0) s, (x_2 = 10) m. Displacement (\Delta x=x_2 - x_1=10 - 4 = 6) m. From graph B (velocity - vs - time graph): The displacement is the area under the velocity - time graph from (t = 3.0) s to (t = 5.0) s. The velocity from (t = 3) s to (t = 4) s is (v_1 = 3) m/s and from (t = 4) s to (t = 5) s is (v_2 = 3) m/s. The area of the rectangle formed by the velocity - time graph from (t = 3) s to (t = 5) s is (A=(4 - 3)\times3+(5 - 4)\times3=3 + 3=6) m.

d.

From graph A (position - vs - time graph): At (t = 7.0) s, (x_1 = 10) m. At (t = 9.0) s, (x_2 = 0) m. Displacement (\Delta x=x_2 - x_1=0 - 10=-10) m. From graph B (velocity - vs - time graph): The velocity from (t = 7) s to (t = 8) s is (v_1=-5) m/s and from (t = 8) s to (t = 9) s is (v_2=-5) m/s. The area of the rectangle formed by the velocity - time graph from (t = 7) s to (t = 9) s is (A=(8 - 7)\times(-5)+(9 - 8)\times(-5)=-5-5=-10) m.

Explanation:

7.

Step1: Initial movement

The object starts at a non - zero position and moves left, so negative slope for position - time graph.

Step2: First stop

When the object stops, position doesn't change with time, so horizontal line.

Step3: Continued left - ward movement

Same direction and speed as before, so same negative slope.

Step4: Second stop

Again, horizontal line as position is constant.

Step5: Right - ward movement

Positive slope, steeper due to greater speed.

8.

a.

Step1: Plot points

Use the data from the (t - x) table to plot points on the graph.

Step2: Connect points

Connect the points with straight - line segments to form the position - time graph.

b.

Step1: Calculate velocities

Use the formula (v=\frac{\Delta x}{\Delta t}) for each time interval.

Step2: Plot velocities

Plot the calculated velocities at the mid - points of the time intervals.

c.

Step1: From position graph

Find positions at (t = 3) s and (t = 5) s and subtract to get displacement.

Step2: From velocity graph

Find the area under the velocity - time graph for the given time interval.

d.

Step1: From position graph

Find positions at (t = 7) s and (t = 9) s and subtract to get displacement.

Step2: From velocity graph

Find the area under the velocity - time graph for the given time interval.