## Solution:

The velocity at time 2.5 seconds is 5 m/s.

The velocity at time 7.5 seconds is -4 m/s.

## Solution:

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

## 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.

## 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.

## 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.

## 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.