College Physics 2.66 – Solving for velocity and acceleration from position graph


Figure 2.68 shows the position graph for a particle for 6 s.

(a) Draw the corresponding Velocity vs. Time graph.

(b) What is the acceleration between 0 s and 2 s?

(c) What happens to the acceleration at exactly 2 s?

position graph for a particle for 6 s.
Figure 2.68

Solution:

Part A

The velocity of the particle is the slope of the position vs time graph. Since the position graph is compose of straight lines, we can say that the velocity is constant for several time ranges.

Time RangesSlope of the position Graph/Velocity
0 to 2 seconds=\frac{2-0}{2-0}=1\:\text{m/s}
2 to 3 seconds=\frac{-3-2}{3-2}=\frac{-5}{1}=-5\:\text{m/s}
3 to 5 seconds0 \text{m/s}
5 to 5 seconds=\frac{-2-\left(-3\right)}{6-5}=\frac{1}{1}=1\:\text{m/s}

Based from the data in the table, we can draw the velocity diagram

velocity vs time graph

Part B

Since the velocity is constant between 0 seconds and 2 seconds, we say that the acceleration is 0.

Part C

Since there is a sudden change in velocity at exactly 2 seconds in a very short amount of time, we say that the acceleration is undefined in this case.


College Physics 2.63 – Constructing a displacement graph


Construct the displacement graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

2.63


Solution:

The position vs time graph is shown in the figure below. 

2.63b

College Physics 2.62 – Acceleration as the slope of the velocity vs time diagram


By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s²  at t = 10 s.

2.62
Figure 2.63

Solution:

Since the graph is a straight line, we can use the two points before and after the specified time to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are \left(0,\:165\right)\:and\:\left(20,\:228\right)

The velocityis computed as

a=m=\frac{\Delta y}{\Delta x}

a=\frac{y_2-y_2}{x_2-x_1}

a=\frac{228\:m/s-165\:m/s}{20\:s-0\:s}

a=3.2\:m/s^2

Therefore, the acceleration is verified to be 3.2 m/s² at t=10 s.


College Physics 2.61 – Velocity as slope of position vs time


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are \left(20,\:6.95\right)\:and\:\left(40,\:11.7\right)

The velocityis computed as

v=m=\frac{\Delta y}{\Delta x}

v=\frac{y_2-y_2}{x_2-x_1}

v=\frac{11.7\:km-6.95\:km}{40\:s-20\:s}

v=0.238\:km/s

Therefore, the velocity is 0.238 km/s.


College Physics 2.60 – Slope of the position vs time diagram


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 10.0 s is 0.208 km/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are \left(0,\:2.80\right)\:and\:\left(20,\:6.95\right)

The velocityis computed as

v=m=\frac{\Delta y}{\Delta x}

v=\frac{y_2-y_2}{x_2-x_1}

v=\frac{6.95\:km-2.80\:km}{20\:s-0\:s}

v=0.208\:km/s

Therefore, the velocity is 0.208 km/s.


College Physics 2.59 – Analysis of motion diagrams


(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is 115 m/s at t = 20 s.

(b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is 5.0 m/s².

Figure 2.60
Figure 2.61

Solution:

Part A

Based from the graph shown in figure 2.60, we can choose two points, one before and one after t=20 s. We have the points \left(15,\:988\right)\:and\:\left(25,\:2138\right).

The slope is computed using the formula:

m=\frac{\Delta y}{\Delta x}

m=\frac{2138\:m-988\:m}{25\:s-15\:s}

m=115\:m/s

Therefore, we have verified that the velocity of the jet car is 115 m/s at t=20 seconds. 

Part B

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are \left(10,\:65\right)\:and\:\left(25,\:140\right)

The acceleration is computed as

a=m=\frac{\Delta y}{\Delta x}

a=\frac{y_2-y_2}{x_2-x_1}

a=\frac{140\:m/s-65\:m/s}{25\:s-10\:s}

a=5\:m/s^2

Therefore, the acceleration is verified to be 5.0 m/s².