Graphical Analysis of One-Dimensional Motion| Kinematics| College Physics| Problem 2.59

(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.

2.59a
Figure 2.60
2.59b
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.

 

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