COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 88P

(a)

To determine

The direction dolphin must head.

(a)

Expert Solution
Check Mark

Answer to Problem 88P

The dolphin must head 30.0°_ north of west.

Explanation of Solution

Consider the positive x direction to be west, and positive y direction to be north.

Figure 1 represents the relative motion of the dolphin with respect to the bay and uniform water current.

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 3, Problem 88P

Write the expression for velocity of dolphin with respect to bay.

    vdb=vdw+vwb        (I)

Here, vdb is the velocity of dolphin with respect to bay, vdw is the velocity of dolphin with respect to the water current, and vwb is the velocity of the water current with respect to the bay.

Write the expression for component of velocity of the dolphin with respect to bay in the x direction.

    vdbx=vdwx+vwbx

Here, vdbx is the velocity of the dolphin with respect to bay in the x direction, vdwx is the velocity of the dolphin with respect to the water current in the x direction, and vwbx is the velocity of the water current with respect to bay in the y direction.

From Figure 1 the above equation is written as

    vdbx=vdwcosθvwbcos45°=vdwcosθvwb2        (II)

Write the expression for component of velocity of the dolphin with respect to bay in the y direction.

    vdby=vdwy+vwby

Here, vdby is the velocity of the dolphin with respect to bay in the y direction, vdwy is the velocity of the dolphin with respect to the water current in the y direction, and vwby is the velocity of the water current with respect to bay in the y direction

From Figure 1 the above equation can be written as

    vdby=vdwsinθvwbsin45°=vdwsinθvwb2        (III)

Since vdby=0, the above is reduced to

    0=vdwsinθvwb2sinθ=vwbvdw2θ=sin1(vwbvdw2)        (IV)

Conclusion:

Substitute 2.83m/s for vwb, and 4.00m/s for vdw in equation (IV), to find θ.

    θ=sin1(2.83m/s(4.00m/s)2)=30.0°

The dolphin should head 30.0° north of west.

Therefore, the dolphin must head 30.0°_ north of west.

(b)

To determine

The time taken by the dolphin to swim 0.80km distant home.

(b)

Expert Solution
Check Mark

Answer to Problem 88P

The time taken by the dolphin to swim 0.80km distant home is 5.47×104s_.

Explanation of Solution

Write the expression for time taken by the dolphin to swim 0.80km distant home.

    Δt=Δxvdbx        (V)

Here, Δt is the time taken by the dolphin to swim 0.80km distant home, Δx is the distance travelled by the dolphin, and vdbx is the component of velocity of the dolphin with respect to bay in the x direction.

From subpart (a), vdbx is derived as vdwcosθvwb2.

Use the above condition in equation (V).

    Δt=Δxvdwcosθvwb2        (VI)

Conclusion:

Substitute 4.00m/s for vdw, 2.83m/s for vwb 0.80km for Δx, and 30.0° for θ in equation (VI), to find Δt.

    Δt=0.80km×1m103km(4.00m/s)cos30.0°2.83m/s2=0.80×103m3.46m/s2.00m/s=5.47×104s

Therefore, the time taken by the dolphin to swim 0.80km distant home is 5.47×104s_.

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Chapter 3 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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