Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 3, Problem 3.43P

Review. As it passes over Grand Bahama Island, the eye of a hurricane is moving in a direction 60.0° north of west with a speed of 41.0 km/h. (a) What is the unit-vector expression for the velocity of the hurricane? It maintains this velocity for 3.00 h. at which time the course of the hurricane suddenly shifts due north, and its speed slows to a constant 25.0 km/h. This new velocity is maintained for 1.50 h. (b) What is the unit-vector expression for the new velocity of the hurricane? (c) What is the unit-vector expression for the displacement of the hurricane during the first 3.00 h? (d) What is the unit-vector expression for the displacement of the hurricane during the latter 1.50 h? (e) How far from Grand Bahama is the eye 4.50 h after it passes over the island?

(a)

Expert Solution
Check Mark
To determine
The unit vector expression for the velocity of hurricane during first 3 hours.

Answer to Problem 3.43P

The unit vector expression for the velocity of hurricane during first 3 hours is (20.5km/h)i^+(35.50km/h)j^ .

Explanation of Solution

Given information:

The velocity of hurricane for the first 3 hours is 41.0km/h with a direction of 60.0° north of west and the velocity of hurricane for next 1.50 hour is 25.0km/h with a direction along north.

Consider the east direction as positive x direction.

The unit vector expression of velocity of hurricane for first 3 hours is given as,

v1=(|v1|cos60.0°)i^+(|v1|sin60.0°)j^ (I)

Substitute 41.0km/h for |v1| in above expression in equation (I).

v1=((41.0km/h)cos60.0°)i^+((41.0km/h)sin60.0°)j^=(20.5km/h)i^+(35.50km/h)j^

Conclusion:

Therefore, the unit vector expression for the velocity of hurricane during first 3 hours is (20.5km/h)i^+(35.50km/h)j^ .

(b)

Expert Solution
Check Mark
To determine
The unit vector expression for the new velocity of the hurricane.

Answer to Problem 3.43P

The unit vector expression for the new velocity of hurricane is (25.0km/h)j^ .

Explanation of Solution

Given information:

The velocity of hurricane for the first 3 hours is 41.0km/h with a direction of 60.0° north of west and the velocity of hurricane for next 1.50 hour is 25.0km/h with a direction along north.

The unit vector expression for the new velocity of the hurricane is,

v2=(|v2|)j^

Substitute 25.0km/h for |v2| in above expression.

v2=(25.0km/h)j^

Conclusion:

Therefore, the unit vector expression for the new velocity of hurricane is (25.0km/h)j^ .

(c)

Expert Solution
Check Mark
To determine
The unit vector expression for the displacement of the hurricane during first 3 hours.

Answer to Problem 3.43P

The unit vector expression for the displacement of the hurricane during first 3 hours is (61.5km)i^+(106.5km)j^ .

Explanation of Solution

Given information:

The velocity of hurricane for the first 3 hours is 41.0km/h with a direction of 60.0° north of west and the velocity of hurricane for next 1.50 hour is 25.0km/h with a direction along north.

Formula to determine the unit vector expression for the displacement during first 3 hours is,

d1=(v1)(t) (II)

  • v1 is the velocity of hurricane during first 3 hours.
  • t is the time.
  • d1 is the displacement of hurricane for 3 hours.

Substitute (20.5km/h)i^+(35.50km/h)j^ for v1 and 3.00hours for t in equation (II).

d1=((20.5km/h)i^+(35.50km/h)j^)(3.00hours)=(61.5km)i^+(106.5km)j^

Conclusion:

Therefore, the unit vector expression for the displacement of the hurricane during first 3 hours is (61.5km)i^+(106.5km)j^ .

(d)

Expert Solution
Check Mark
To determine
The unit vector expression for the displacement of the hurricane during the later 1.50 hours.

Answer to Problem 3.43P

The unit vector expression for the displacement of the hurricane during the later 1.50 hours is (37.5km)j^ .

Explanation of Solution

Given information:

The velocity of hurricane for the first 3 hours is 41.0km/h with a direction of 60.0° north of west and the velocity of hurricane for next 1.50 hour is 25.0km/h with a direction along north.

Formula to determine the unit vector expression for the displacement during the later 1.50 hours is,

d2=(v2)(t) (III)

  • v1 is the velocity of hurricane during the later 1.50 hours.
  • t is the time.
  • d1 is the displacement of hurricane for the later 1.50 hours.

Substitute (25.0km/h)j^ for v2 and 1.50hours for t in equation (III).

d1=((25.0km/h)j^)(1.50hours)=(37.5km)j^

Conclusion:

Therefore, the unit vector expression for the displacement of the hurricane during the later 1.50 hours is (37.5km)j^ .

(e)

Expert Solution
Check Mark
To determine
The net displacement of the hurricane after 4.50hours .

Answer to Problem 3.43P

The net displacement of the hurricane after 4.50hours is 157km .

Explanation of Solution

Section 1:

To determine: The unit vector expression for the net displacement after 4.50hours .

Answer: The unit vector expression for the net displacement after 4.50hours is (61.5km)i^+(144km)j^ .

Explanation:

Given information:

The velocity of hurricane for the first 3 hours is 41.0km/h with a direction of 60.0° north of west and the velocity of hurricane for next 1.50 hour is 25.0km/h with a direction along north.

Formula to determine the unit vector expression for the net displacement after 4.50hours is,

d=d1+d2 (IV)

Substitute (61.5km)i^+(106.5km)j^ for d1 and (37.5km)j^ for d2 in equation (IV).

d=(61.5km)i^+(106.5km)j^+(37.5km)j^=(61.5km)i^+(144km)j^

Section 2:

To determine: The magnitude of the net displacement after 4.50hours .

Answer: The magnitude of the net displacement after 4.50hours is 157km .

Explanation:

Given information:

The velocity of hurricane for the first 3 hours is 41.0km/h with a direction of 60.0° north of west and the velocity of hurricane for next 1.50 hour is 25.0km/h with a direction along north.

From part (d), the unit vector expression for the net displacement after 4.50hours is given as,

d=(61.5km)i^+(144km)j^

Formula to calculate the magnitude of the net displacement after 4.50hours is,

|d|=(dx)2+(dy)2 (V)

Substitute 61.5km for dx and 144km for dy in equation (V).

|d|=(61.5km)2+(144km)2=156.58km157km

Conclusion:

Therefore, the magnitude of the net displacement after 4.50hours is 157km

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Chapter 3 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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