PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 42P

(a)

To determine

The average speed of the trip.

(a)

Expert Solution
Check Mark

Answer to Problem 42P

The average speed of the trip is 102km/h.

Explanation of Solution

The speed of the speed boat for first 20.0min is 108km/h. The speed of the boat for next 10.0min is 90.0km/h.

Write the formula for the average speed of the speed boat.

va=v1t1+v2t2t1+t2 (I)

Here, va is the average speed, v1 is the first velocity, t1 is the time for which the speed is v1, v2 is the second velocity, t2 is the time for which the speed is v2

Conclusion:

Substitute 20.0min for t1, 10.0min for t2, 108km/h for v1, 90.0km/h for v2 in equation (I).

va=(108km/h)(20.0min)+(90.0km/h)(10.0min)20.0min+10.0min=102km/h

The average speed of the trip is 102km/h.

(b)

To determine

The average velocity of the trip.

(b)

Expert Solution
Check Mark

Answer to Problem 42P

The average velocity is 90.8km/h and directed 16.6° south of the west direction.

Explanation of Solution

The speed of the speed boat for first 20.0min is 108km/h towards west. The speed of the boat for next 10.0min is 90.0km/h towards 60.0° south of west.

Write the distance covered in west direction.

d1=v1t1 (I)

Here, d1 is the distance covered in west direction, v1 is the first velocity, t1 is the time for which the speed is v1.

Write the distance covered in the direction towards 60.0° south of west.

d2=v2t2 (II)

Here, d2 is the distance covered in the direction towards 60.0° south of west, v2 is the second velocity, t2 is the time for which the speed is v2.

The figure below shows the entire trip.

PHYSICS, Chapter 3, Problem 42P

Write the formula for the x-component of the displacement.

dx=d1+d2cos240.0° (III)

Here, dx is the x-component of the displacement.

Write the formula for the y-component of the displacement.

dy=d2sin240.0° (IV)

Here, dy is the y-component of the displacement.

Write the formula for the magnitude of the displacement.

|d|=dx2+dy2 (V)

Here, d is the displacement.

Write the formula for the angle made by the displacement vector.

θ=tan1(dydx) (VI)

Here, θ is the angle made by the velocity vector.

Write the formula for the average velocity.

|v|=dt (VII)

Here, v is the velocity vector, t is the total time for the trip.

Conclusion:

Substitute 20.0min for t1, 108km/h for v1 in equation (I).

d1=(108km/h)(20.0min)=(108km/h)(20.0min)(1h60min)=36.0km

Substitute 10.0min for t2, 90.0km/h for v2 in equation (II).

d2=(90.0km/h)(10.0min)=(90.0km/h)(10.0min)(1h60min)=15.0km

Substitute 36.0km for d1, 15.0km for d2 in equation (III).

dx=36.0km+(15.0km)cos240.0°=43.5km

Substitute 15.0km for d2 in equation (IV).

dy=(15.0km)sin240.0°=13.0km

Substitute 43.5km for dx, 13.0km for dy in equation (V).

|d|=(43.5km)2+(13.0km)2=45.4km

Substitute 43.5km for dx, 13.0km for dy in equation (VI).

θ=tan1(13.0km43.5km)=16.6°

Substitute 45.4km for d, 30.0min for t in equation (VII).

|v|=45.4km30.0min=45.4km(30.0min)(1h60min)=90.8km/h

The average velocity is 90.8km/h and directed 16.6° south of the west direction.

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Chapter 3 Solutions

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