Question

1. 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

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.

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.