COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 3, Problem 123P

(a)

To determine

The direction and distance the person should travel to reach the starting point.

(a)

Expert Solution
Check Mark

Answer to Problem 123P

The person has to walk 6.3km at 29.6° angle with horizontal to reach the starting point.

Explanation of Solution

Draw the vector diagram showing the displacements of the person.

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 3, Problem 123P

Represent the displacements of person as vectors. Let west direction is taken as -ve x-direction, east direction as +ve x-direction, north as +y-axis and south as -ve y-axis.

Express 2.00 km in west direction as a vector.

    A=(2.00km)i^

Express 5.00 km in south-west direction as a vector.

    B=[(5.00km)(cos53°)]i^[(5.00km)(sin53°)]j^=(3.0km)i^(4.0km)j^

Express 1.00 km in north-west direction as a vector.

    C=[(1.00km)(cos60°)]i^+[(1.00km)(sin60°)]j^=(0.5km)i^+(0.87km)j^

The net displacement is given by the vector addition.

    R=A+B+C

Here, R is the net displacement vector.

Rewrite the above equation by substituting the equations for A, B and C.

  R=(2.00km)i^+(3.0km)i^(4.0km)j^+(0.5km)i^+(0.87km)j^=(5.5km)i^(3.13km)j^

Write the equation to find the magnitude of R.

    |R|=x2+y2

Here, |R| is the magnitude, x is the x-component and y is the y-component.

Write the equation to find the direction of R.

    θ=tan1(yx)

Here, θ is the direction.

Conclusion:

Substitute 5.5km for x and 3.13km for y in the equation for |R|.

    |R|=(5.5km)2+(3.13km)2=40.05km2=6.3km

Substitute 5.5km for x and 3.13km for y in the equation for θ.

    θ=tan1(3.13km5.5km)=tan1(0.569)=29.6°

Thus, the person has to walk 6.3km at 29.6° angle with horizontal to reach the starting point.

(b)

To determine

The time taken by the person to reach starting from the final point at the speed of 5.00m/s.

(b)

Expert Solution
Check Mark

Answer to Problem 123P

Time taken will be 21min.

Explanation of Solution

Write the equation to find the time taken.

    t=dv

Here, t is the time taken and d is the net displacement.

Conclusion:

Substitute 6.3km for d and 5.00m/s for v in the above expression.

    t=6.3km(103m1km)5.00m/s=1.26×103s(1min60s)=21min

Here, the time taken will be 21min.

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Chapter 3 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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