#### Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.59.

## Solution:

**North Component**

The North component is given by

**East Component**

The east component is

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# Tag: Displacement

#### Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.59.

## Solution:

**North Component**

**East Component**

#### (a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R′ =A−B ).

#### (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R′′ = B – A = – R′ ). Show that this is the case.

**Solution:**

**Part A**

**Part B**

#### Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B , which is 20.0 m in a direction exactly 40º south of west, and then leg A , which is 12.0 m in a direction exactly 20º west of north. (This problem shows that A+B=B+A.)

**Solution:**

#### Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.54, then this problem finds their sum R=A+B.)

**Solution:**

#### Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.53, then this problem asks you to find their sum R = A + B .)

**Solution:**

#### Find the north and east components of the displacement for the hikers shown in Figure 3.50.

**Solution:**

#### Find the following for path B in Figure 3.52:

#### (a) the total distance traveled, and

#### (b) the magnitude and direction of the displacement from start to finish.

**Solution:**

**Part A**

**Part B**

#### Find the following for path A in Figure 3.52:

#### (a) the total distance traveled, and

#### (b) the magnitude and direction of the displacement from start to finish.

**Solution:**

**Part A**

**Part B**

#### Freight trains can produce only relatively small accelerations and decelerations.

#### (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s^{2} for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

#### (b) If the train can slow down at a rate of 0.550 m/s^{2}, how long will it take to come to a stop from this velocity?

#### (c) How far will it travel in each case?

**Solution:**

### Part A

**Part B**

### Part C

**Find the following for path D in the figure: **

**(a) The distance traveled. **

**(b) The magnitude of the displacement from start to finish. **

**(c) The displacement from start to finish. **

Continue reading “College Physics Problem 2.4”

The North component is given by

The east component is

We are required to solve for

The x-component of the resultant is:

The y-component of the resultant is:

The resultant is

The compass direction is

The compass direction is

The result is the same as that in Part A since the components did not change. Only the order has changed. This is consistent with the fact that

Reversing the order of the vectors, we come up with the triangle as shown below.

We can actually see that the resultant can be solved by considering the triangle formed by the three vectors. The same with the procedures used in the previous problem, we have

To solve for the direction, we can solve for the angle included between B and R first.

Finally to solve for the direction of the resultant, we have

The compass direction is .

This result is actually the same as in the previous problem when the order of the vectors are not reversed. No matter what is the order of the vectors we are adding, the result is still the same. This proves that .

To solve for the magnitude and direction of the resultant R, we can refer to the triangle formed by the tree vectors as shown in the figure below.

From the figure above, we are given the two sides A and B, and the included angle measured to be 70°. To solve for the magnitude of R, we can use the cosine law.

To solve for the direction of the resultant, we just need to solve for angle C, then subtract 70° to get the value of θ. We can solve for angle C using the sine law.

Finally, to solve for θ, we subtract 70° from 74.65°.

Therefore, the compass direction of the resultant displacement is 4.65° South of West.

Refer to the diagram

The magnitude of the displacement (distance from the starting point to the final position) is given by

The compass direction of a line connecting your starting point to your final position is given by

The compass reading is .

To solve for the north and east components of the given displacement vector, we need to draw them first by virtue of the parallelogram law. The vector, together with its components, is shown in the figure below.

We can solve the north and east components by considering the right triangle ABC. The east component is the side AC and the north component is the side BC.

The North-component is given by

The East-component is given by

So, the total distance traveled on path B as shown by the red line is just the sum of the distances.

The displacement is just the shortest distance from the beginning to the end of the path. The displacement has a magnitude given by

The direction of the displacement is given by

So, the total distance traveled on path A as shown by the green line is just the sum of the distances.

The displacement is just the shortest distance from the beginning to the end of the path. The displacement has a magnitude given by

The direction of the displacement is given by

The final velocity can be solved using the formula . We substitute the given values.

Rearrange the equation we used in part (a) by solving in terms of t, we have

The change in position for part (a), , or distance traveled is computed using the formula

For the situation in part (b), the distance traveled is computed using the formula

Continue reading “College Physics Problem 2.4”

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