Question

A disoriented physics professor drives 3.35 km north, then 4.67 km west, and then 1.69 km south.

Part A

Find the magnitude of the resultant displacement of this professor.

Express your answer in kilometers.

Part B

Find the direction of the resultant displacement of this professor.

Express your answer in degrees.

Answer 1

Given: A professor goes 3.35 Km North, 4.67 Km West, and then 1.69 Km South.

Solution:

(a)

Fig. 1 represents the problem statement. The North, West, South, East directions are taken as shown in the diagram.

To find the resultant displacement, we use the triangle law of vector addition. For this, we break the diagram into two triangles, OAB and OCB as shown in Fig. 2

Fig. 2

Finding out the resultant OB using triangle law,

Fig. 3. The first diagram represents the resultant OB and the second diagram represents resultant OC.

Let vectors OA= 3.35, AB= 4.67 and vector OB is to be found out.

Apllying triangle law,

Thus from OAB, the value of can be found out by

Using the property of sum of angle on a straight line is , the value of is

Therefore, from the second diagram in Fig. 3, vectors OB=5.74, BC=1.69 and the resultant OC is to be found out.

Applying Triangle law of vector addition again,

Answer: The value of the resultant displacement is 4.949 Km from the origin.

(b)

Let the angle in Fig. 3 second diagram be

Then is given by

Therefore, from Fig 2, the value of angle

The values of and are already found out above. Putting in the values we get

Answer: The direction of the resultant vector is 70.40 degrees counter clockwise from positive X-axis (Direction of North pole).

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