# 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.

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.