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Get Full Access to University Physics - 13 Edition - Chapter 1 - Problem 34e
Get Full Access to University Physics - 13 Edition - Chapter 1 - Problem 34e

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# A postal employee drives a delivery truck over the route

ISBN: 9780321675460 31

## Solution for problem 34E Chapter 1

University Physics | 13th Edition

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Problem 34E

A postal employee drives a delivery truck over the route shown in Fig. E1.25. Use the method of components to determine the magnitude and direction of her resultant displacement. In a vector-addition diagram (roughly to scale), show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained by using the method of components.

Step-by-Step Solution:

Solution 34E Consider this as the path travelled by the postal employee. Since she is moving towards the right top corner, we can infer her direction is towards north-east. Step 1: According to the law of vector addition, we can consider, AB + BC = AC 2 2 So, the magnitude of vector AC will be, A| = |B +BC + 2AB×BC cos Magnitude of AB = 2.6 km Magnitude of BC = 4 km Angle between the vectors, = 90° 2 2 2 Therefore, |AC | (2.6 km) + (4km) = 6.76 + 16 = 22.6 km = 4.77 km We know that, AC + CD = AD Vector AD is the resultant of vectors what we need and the magnitude will give us the total displacement of the postal employee from A to D. Provided, magnitude of CD = 3.1 km in the diagram Therefore, |D | A + CD + 2AC × CD cos = DCA = DCB + ACB DCB = 180° - DCF DCF = 45° DCB = 180° - 45° = 135° From the diagram, tan ( ACB) = 2.6 km / 4 km = 0.65 ACB = tan (0.65) = 33° = DCA = DCB + ACB = 135° + 33° = 168° Putting this back in the equation for magnitude. 2 2 |AD | 22.76 km + (3.1km) + 2 × 4.77 × 3.1 × cos (180 168) Cos 12 = 0.9781 We should take the angle value as (180 - ) if it is more than 90°. |AD | 22.76 km + 9.61km + (2 × 4.77 × 3.1 × 0.9781) = 32.37 + 28.93 = 61. km 2 So, the displacement for the postal employee from starting point till ending point will be 7.83 km.

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