# Graphical Analysis of One-Dimensional Motion| Kinematics| College Physics| Problem 2.62

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

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.