PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 79P

(a)

To determine

The displacement of the plane from its starting point.

(a)

Expert Solution
Check Mark

Answer to Problem 79P

The person S’s speed with respect to the starting point on the bank is 1.80mi/h .

Explanation of Solution

The person S’s motion is shown in Figure 1.

PHYSICS, Chapter 3, Problem 79P

Write an expression to calculate the x component of the speed of person S with respect to the starting point on the bank.

vSbx=vcosθ+vx (I)

Here, vSbx is the x component of the speed of person S with respect to the starting point on the bank, v is the speed of the person s, θ is the angle and vx is the x component of the speed of stream.

Write an expression to calculate the y component of the speed of the person S.

vSby=vsinθ+vy (II)

Here, vSby is the y component of the speed of person S with respect to the starting point on the bank and vy is the y component of the speed of stream.

Write an expression to calculate the speed of the person with respect to the starting point on the bank.

vSb=vSbx2+vSby2 (III)

Conclusion:

Substitute 3.00mi/h for v, 0mi/h for vx , and 60.0° for θ  in equation (I) to find vSbx.

vSbx=(3.00mi/h)cos(60.0°)+0mi/h=(3.00mi/h)(12)=1.50mi/h

Substitute 3.00mi/h for v, 1.60mi/h for vy , and 60.0° for θ  in equation (II) to find vSby.

vSby=(3.00mi/h)sin(60.0°)+1.60mi/h=(3.00mi/h)(32)1.60mi/h=1.00mi/h

Substitute 1.50mi/h for vSbx, and 1.00mi/h for vSby in equation (III) to find vSb.

vSb=(1.50mi/h)2+(1.00mi/h)2=3.25mi/h=1.80mi/h

Thus, the person S’s speed with respect to the starting point on the bank is 1.80mi/h .

(b)

To determine

The direction at which the jetliner have flown directly to the same destination.

(b)

Expert Solution
Check Mark

Answer to Problem 79P

The time taken to cross the river is 48.0min.

Explanation of Solution

Write an expression to calculate the time taken to cross the river.

Δt=Δxvcosθ (IV)

Here, Δt is the time taken to cross the river and Δx is the width of the river.

Conclusion:

Substitute 1.20mi for Δx, 3.00mi/h for v, and 60.0° for θ in equation (IV) to find Δt.

Δt=1.20mi(3.00mi/h)cos(60.0°)=1.20mi(3.00mi/h)cos(60.0°)=1.20mi(3.00mi/h)(12)=(0.800h)(60min1hr)=48.0min

Thus, the time taken to cross the river is 48.0min.

(c)

To determine

The time taken for the trip.

(c)

Expert Solution
Check Mark

Answer to Problem 79P

The upstream or downstream of the person S from the starting point the person S reaches at the opposite bank is 0.800miupstream.

Explanation of Solution

Write an expression to calculate the upstream or downstream of the person S from the starting point the person S reaches at the opposite bank.

Δy=vsinθΔt (V)

Here, Δy is the upstream or downstream of the person S from the starting point the person S reaches at the opposite bank.

Conclusion:

Substitute 0.800h for Δt, 3.00mi/h for v, and 60.0° for θ in equation (V) to find Δy.

Δy=(3.00mi/h)sin(60.0°)(0.800h)=(3.00mi/h)(32)(0.800h)=0.800miupstream

Thus, the upstream or downstream of the person S from the starting point the person S reaches at the opposite bank is 0.800miupstream.

(d)

To determine

The time taken for the direct flight.

(d)

Expert Solution
Check Mark

Answer to Problem 79P

The upstream angle the person S should go in order to go straight across is 32.2°upstream.

Explanation of Solution

The upstream component of her velocity relative to the water should be equal in magnitude to the velocity of the current relative to the bank.

Write an expression to calculate the launch angle θ at which the maximum range occurs.vSby=0vsinθ'+vy=0 (VI)

Here, θ' is upstream angle the person S should go in order to go straight across.

Rearrange the equation (VI) to find θ'.

θ'=sin1(vyv) (VII)

Conclusion:

Substitute 3.00mi/h for v, and 1.60mi/h for vy   in equation (VII) to find vSby.

θ'=sin1((1.60mi/h)3.00mi/h)=sin1(1.60mi/h3.00mi/h)=32.2°upstream

Thus, the upstream angle the person S should go in order to go straight across is 32.2°upstream.

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Chapter 3 Solutions

PHYSICS

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