## Solution:

So, we are given the two vectors shown below.

If we use the graphical method of adding vectors, we can join the two vectors using head-tail addition and come up with the following:

The resultant is drawn from the tail of the first vectors (the origin) to the head of the last vector. The resultant is shown in red in the figure below. Figure 3.9B: The resultant of vectors A and B using graphical method

The result is in conformity with that in figure 3.24 shown on the question shown above.

## Solution:

Consider the three vectors shown in the figure below:

First, we shall add them A+B+C. Using the head-tail or graphical method of vector addition, we have the figure shown below.

Now, let us try to find the sum of the three vectors by reordering vectors A, B, and C. Let us try to find the sum of C+B+A in that order. The result is shown below.

We can see that the resultant is the same directed from the origin upward. This proves that the resultant must be the same even if the vectors are added in different order.

## Solution:

### Part A

The velocity of the particle is the slope of the position vs time graph. Since the position graph is compose of straight lines, we can say that the velocity is constant for several time ranges.

Based from the data in the table, we can draw the velocity diagram

### Part B

Since the velocity is constant between 0 seconds and 2 seconds, we say that the acceleration is 0.

### Part C

Since there is a sudden change in velocity at exactly 2 seconds in a very short amount of time, we say that the acceleration is undefined in this case.

## Solution:

### Part A

To find for the average velocity over the straight line graph of the velocity vs time shown, we just need to locate the midpoint of the line. In this case, the average speed for the first 4 seconds is $\text{v}_{\text{ave}}=6\:\text{m/s}$

### Part B

Looking at the graph, the velocity at exactly 5 seconds is 12 m/s.

### Part C

If we are given the velocity-time graph, we can solve for the acceleration by solving for the slope of the line.

Consider the line from time zero to time, t=4 seconds. The slope, or acceleration, is $\text{a}=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2$

### Part D

For the first 4 seconds, the distance traveled is equal to the area under the curve. $\text{distance}=\frac{1}{2}\left(4\:\sec \right)\left(12\:\text{m/s}\right)=24\:\text{m}$

So, the sprinter traveled a total of 24 meters in the first 4 seconds. He still needs to travel a distance of 76 meters to cover the total racing distance. At the constant rate of 12 m/s, he can run the remaining distance by $\text{t}=\frac{\text{distance}}{\text{velocity}}=\frac{76\:\text{m}}{12\:\text{m/s}}=6.3\:\sec$

Therefore, the total time of the sprint is $\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec$