## 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.

## Find Acceleration Diagram from Velocity Vector Diagram| University Physics

### The figure above shows the motion diagram for an object. If the x-axis is chosen as shown (positive to the right), which of the diagrams below best represents the corresponding acceleration vs time graph? (In the graph, later time is to the right). ## Pendulum’s acceleration at the lowest point| University Physics

### In a physics lab, a pendulum hung from the ceiling slowly swings back and forth. Select the arrow that best indicates the direction of the pendulum’s acceleration when it reaches the lowest point moving from left to right. ## Rank the diagrams according to the x-component of the acceleration| University Physics

### The figure below shows six different motion diagrams. The separation between dots is in the same scale in all diagrams, and the time interval between dots is also the same for all of them. The direction of the positive x-axis is also shown. Rank the diagrams according to the x-component of the acceleration. (Positive is greater than negative) 