Knowing that the coordinates of the utility pole are x p = 46 ft and y p = 59 ft, determine ( a ) the time elapsed from the first collision to the stop at P , ( b ) the speed of car A .

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Answer 1

Textbook Question

Chapter 14.1, Problem 14.20P

Knowing that the coordinates of the utility pole are x p = 46 ft and y p = 59 ft, determine (a) the time elapsed from the first collision to the stop at P, (b) the speed of car A.

Expert Solution

To determine

(a)

Time elapsed from first collision to the stop at P.

Answer to Problem 14.20P

The time elapsed from first collision to stop at pole is 2.61 s.

Explanation of Solution

Given:

Speed of car B is 45 mi/hr.

Speed of car C is 60 mi/hr.

Weight if car A is 3000 lb.

Weight of car B is 2600 lb.

Weight of car C is 2400 lb.

Coordinates of utility pole are (46 ft,59 ft).

Concept used:

Draw the diagram for the system mentioned as shown below:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.20P

Assume origin as point O where, Car A and Car B collides.

Write the expression for the mass center of first collision.

(mA+mB+mC)(x0i^+y0j^)=mA(xAi^+yAj^)+mB(xBi^+yBj^)+mC(xCi^+yCj^) ...... (1)

Multiply the whole equation (1) with g on both sides.

(WA+WB+WC)(x0i^+y0j^)=WA(xAi^+yAj^)+WB(xBi^+yBj^)+WC(xCi^+yCj^) ...... (2)

Here, WA is the weight of car A, WB is the weight of car B, WC is the weight of car C, (x0,y0) is coordinate of center of mass, (xA,yA) is the coordinate of car A, (xB,yB) is the coordinate of car B and (xC,yC) is the coordinate of car C.

Write the expression for conservation of momentum for the system.

(mA+mB+mC)v→=mAv→A+mBv→B+mCv→C ...... (3)

Multiply the equation (3) with g on both sides.

(WA+WB+WC)v→=WAv→A+WBv→B+WCv→C ...... (4)

Here, v→A is the velocity of car A, v→B is the velocity of car B, v→C is the velocity of car C and v→ is the velocity of center of mass of system.

Write the expression for the position of pole P.

xPi^+ypj^=x0i^+y0j^+v→t ...... (5)

Here, (xP,yP) is the coordinate of pole P.

Calculation:

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, 0 for xA, 0 for yA, 0 for xB, 0 for yB, 32 ft for xC and 10 ft for yC in equation (2).

(3000+2600+2400)(x0i^+y0j^)=[3000(0i^+0j^)+2600(0i^+0j^)+2400(32i^+10j^)]8000x0i^+8000y0j^=76800i^+24000j^

Equate i^ components of above expression.

x0=9.6 ft

Equate j^ components of the above expression.

y0=3 ft

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, −(60 mi/hr)i^ for v→C and (45 mi/hr)j^ for v→B in equation (4).

(3000+2600+2400)v→=[3000v→A+2600(45 mi/hr(1.467 ft/s1 mi/hr))j^+2400(−60 mi/hr(1.467 ft/s1 mi/hr))i^]8000v→=3000v→A+171639j^−211248i^v→=0.375v→A+21.455j^−26.406i^

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate j^ components of above expression.

59=(3+21.455t)56=21.455t

Rearrange the above expression for t.

t=2.61 s

The time elapsed from first collision to stop at pole is 2.61 s.

Conclusion:

Thus, the time elapsed from first collision to stop at pole is 2.61 s.

Expert Solution

To determine

(b)

Speed of car A.

Answer to Problem 14.20P

The velocity of car A is 73.3 mi/hr.

Explanation of Solution

Calculation:

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate the i^ components of both side in above expression.

46=[9.6+(0.375vA−26.406)t]36.4=(0.375vA−26.406)t

Substitute 2.61 s for t in above expression.

36.4=(0.375vA−26.406)(2.61)36.4=0.97875vA−68.92

Simplify above expression for vA.

vA=107.6 ft/s(0.681 mi/hr1 ft/s)=73.3 mi/hr

The velocity of car A is 73.3 mi/hr.

Conclusion:

Thus, the velocity of car A is 73.3 mi/hr.

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Answer 2

Textbook Question

Chapter 14.1, Problem 14.20P

Knowing that the coordinates of the utility pole are x p = 46 ft and y p = 59 ft, determine (a) the time elapsed from the first collision to the stop at P, (b) the speed of car A.

Expert Solution

To determine

(a)

Time elapsed from first collision to the stop at P.

Answer to Problem 14.20P

The time elapsed from first collision to stop at pole is 2.61 s.

Explanation of Solution

Given:

Speed of car B is 45 mi/hr.

Speed of car C is 60 mi/hr.

Weight if car A is 3000 lb.

Weight of car B is 2600 lb.

Weight of car C is 2400 lb.

Coordinates of utility pole are (46 ft,59 ft).

Concept used:

Draw the diagram for the system mentioned as shown below:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.20P

Assume origin as point O where, Car A and Car B collides.

Write the expression for the mass center of first collision.

(mA+mB+mC)(x0i^+y0j^)=mA(xAi^+yAj^)+mB(xBi^+yBj^)+mC(xCi^+yCj^) ...... (1)

Multiply the whole equation (1) with g on both sides.

(WA+WB+WC)(x0i^+y0j^)=WA(xAi^+yAj^)+WB(xBi^+yBj^)+WC(xCi^+yCj^) ...... (2)

Here, WA is the weight of car A, WB is the weight of car B, WC is the weight of car C, (x0,y0) is coordinate of center of mass, (xA,yA) is the coordinate of car A, (xB,yB) is the coordinate of car B and (xC,yC) is the coordinate of car C.

Write the expression for conservation of momentum for the system.

(mA+mB+mC)v→=mAv→A+mBv→B+mCv→C ...... (3)

Multiply the equation (3) with g on both sides.

(WA+WB+WC)v→=WAv→A+WBv→B+WCv→C ...... (4)

Here, v→A is the velocity of car A, v→B is the velocity of car B, v→C is the velocity of car C and v→ is the velocity of center of mass of system.

Write the expression for the position of pole P.

xPi^+ypj^=x0i^+y0j^+v→t ...... (5)

Here, (xP,yP) is the coordinate of pole P.

Calculation:

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, 0 for xA, 0 for yA, 0 for xB, 0 for yB, 32 ft for xC and 10 ft for yC in equation (2).

(3000+2600+2400)(x0i^+y0j^)=[3000(0i^+0j^)+2600(0i^+0j^)+2400(32i^+10j^)]8000x0i^+8000y0j^=76800i^+24000j^

Equate i^ components of above expression.

x0=9.6 ft

Equate j^ components of the above expression.

y0=3 ft

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, −(60 mi/hr)i^ for v→C and (45 mi/hr)j^ for v→B in equation (4).

(3000+2600+2400)v→=[3000v→A+2600(45 mi/hr(1.467 ft/s1 mi/hr))j^+2400(−60 mi/hr(1.467 ft/s1 mi/hr))i^]8000v→=3000v→A+171639j^−211248i^v→=0.375v→A+21.455j^−26.406i^

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate j^ components of above expression.

59=(3+21.455t)56=21.455t

Rearrange the above expression for t.

t=2.61 s

The time elapsed from first collision to stop at pole is 2.61 s.

Conclusion:

Thus, the time elapsed from first collision to stop at pole is 2.61 s.

Expert Solution

To determine

(b)

Speed of car A.

Answer to Problem 14.20P

The velocity of car A is 73.3 mi/hr.

Explanation of Solution

Calculation:

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate the i^ components of both side in above expression.

46=[9.6+(0.375vA−26.406)t]36.4=(0.375vA−26.406)t

Substitute 2.61 s for t in above expression.

36.4=(0.375vA−26.406)(2.61)36.4=0.97875vA−68.92

Simplify above expression for vA.

vA=107.6 ft/s(0.681 mi/hr1 ft/s)=73.3 mi/hr

The velocity of car A is 73.3 mi/hr.

Conclusion:

Thus, the velocity of car A is 73.3 mi/hr.

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Answer 3

Textbook Question

Chapter 14.1, Problem 14.20P

Knowing that the coordinates of the utility pole are x p = 46 ft and y p = 59 ft, determine (a) the time elapsed from the first collision to the stop at P, (b) the speed of car A.

Expert Solution

To determine

(a)

Time elapsed from first collision to the stop at P.

Answer to Problem 14.20P

The time elapsed from first collision to stop at pole is 2.61 s.

Explanation of Solution

Given:

Speed of car B is 45 mi/hr.

Speed of car C is 60 mi/hr.

Weight if car A is 3000 lb.

Weight of car B is 2600 lb.

Weight of car C is 2400 lb.

Coordinates of utility pole are (46 ft,59 ft).

Concept used:

Draw the diagram for the system mentioned as shown below:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.20P

Assume origin as point O where, Car A and Car B collides.

Write the expression for the mass center of first collision.

(mA+mB+mC)(x0i^+y0j^)=mA(xAi^+yAj^)+mB(xBi^+yBj^)+mC(xCi^+yCj^) ...... (1)

Multiply the whole equation (1) with g on both sides.

(WA+WB+WC)(x0i^+y0j^)=WA(xAi^+yAj^)+WB(xBi^+yBj^)+WC(xCi^+yCj^) ...... (2)

Here, WA is the weight of car A, WB is the weight of car B, WC is the weight of car C, (x0,y0) is coordinate of center of mass, (xA,yA) is the coordinate of car A, (xB,yB) is the coordinate of car B and (xC,yC) is the coordinate of car C.

Write the expression for conservation of momentum for the system.

(mA+mB+mC)v→=mAv→A+mBv→B+mCv→C ...... (3)

Multiply the equation (3) with g on both sides.

(WA+WB+WC)v→=WAv→A+WBv→B+WCv→C ...... (4)

Here, v→A is the velocity of car A, v→B is the velocity of car B, v→C is the velocity of car C and v→ is the velocity of center of mass of system.

Write the expression for the position of pole P.

xPi^+ypj^=x0i^+y0j^+v→t ...... (5)

Here, (xP,yP) is the coordinate of pole P.

Calculation:

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, 0 for xA, 0 for yA, 0 for xB, 0 for yB, 32 ft for xC and 10 ft for yC in equation (2).

(3000+2600+2400)(x0i^+y0j^)=[3000(0i^+0j^)+2600(0i^+0j^)+2400(32i^+10j^)]8000x0i^+8000y0j^=76800i^+24000j^

Equate i^ components of above expression.

x0=9.6 ft

Equate j^ components of the above expression.

y0=3 ft

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, −(60 mi/hr)i^ for v→C and (45 mi/hr)j^ for v→B in equation (4).

(3000+2600+2400)v→=[3000v→A+2600(45 mi/hr(1.467 ft/s1 mi/hr))j^+2400(−60 mi/hr(1.467 ft/s1 mi/hr))i^]8000v→=3000v→A+171639j^−211248i^v→=0.375v→A+21.455j^−26.406i^

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate j^ components of above expression.

59=(3+21.455t)56=21.455t

Rearrange the above expression for t.

t=2.61 s

The time elapsed from first collision to stop at pole is 2.61 s.

Conclusion:

Thus, the time elapsed from first collision to stop at pole is 2.61 s.

Expert Solution

To determine

(b)

Speed of car A.

Answer to Problem 14.20P

The velocity of car A is 73.3 mi/hr.

Explanation of Solution

Calculation:

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate the i^ components of both side in above expression.

46=[9.6+(0.375vA−26.406)t]36.4=(0.375vA−26.406)t

Substitute 2.61 s for t in above expression.

36.4=(0.375vA−26.406)(2.61)36.4=0.97875vA−68.92

Simplify above expression for vA.

vA=107.6 ft/s(0.681 mi/hr1 ft/s)=73.3 mi/hr

The velocity of car A is 73.3 mi/hr.

Conclusion:

Thus, the velocity of car A is 73.3 mi/hr.

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Answer 4

Textbook Question

Chapter 14.1, Problem 14.20P

Knowing that the coordinates of the utility pole are x p = 46 ft and y p = 59 ft, determine (a) the time elapsed from the first collision to the stop at P, (b) the speed of car A.

Expert Solution

To determine

(a)

Time elapsed from first collision to the stop at P.

Answer to Problem 14.20P

The time elapsed from first collision to stop at pole is 2.61 s.

Explanation of Solution

Given:

Speed of car B is 45 mi/hr.

Speed of car C is 60 mi/hr.

Weight if car A is 3000 lb.

Weight of car B is 2600 lb.

Weight of car C is 2400 lb.

Coordinates of utility pole are (46 ft,59 ft).

Concept used:

Draw the diagram for the system mentioned as shown below:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.20P

Assume origin as point O where, Car A and Car B collides.

Write the expression for the mass center of first collision.

(mA+mB+mC)(x0i^+y0j^)=mA(xAi^+yAj^)+mB(xBi^+yBj^)+mC(xCi^+yCj^) ...... (1)

Multiply the whole equation (1) with g on both sides.

(WA+WB+WC)(x0i^+y0j^)=WA(xAi^+yAj^)+WB(xBi^+yBj^)+WC(xCi^+yCj^) ...... (2)

Here, WA is the weight of car A, WB is the weight of car B, WC is the weight of car C, (x0,y0) is coordinate of center of mass, (xA,yA) is the coordinate of car A, (xB,yB) is the coordinate of car B and (xC,yC) is the coordinate of car C.

Write the expression for conservation of momentum for the system.

(mA+mB+mC)v→=mAv→A+mBv→B+mCv→C ...... (3)

Multiply the equation (3) with g on both sides.

(WA+WB+WC)v→=WAv→A+WBv→B+WCv→C ...... (4)

Here, v→A is the velocity of car A, v→B is the velocity of car B, v→C is the velocity of car C and v→ is the velocity of center of mass of system.

Write the expression for the position of pole P.

xPi^+ypj^=x0i^+y0j^+v→t ...... (5)

Here, (xP,yP) is the coordinate of pole P.

Calculation:

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, 0 for xA, 0 for yA, 0 for xB, 0 for yB, 32 ft for xC and 10 ft for yC in equation (2).

(3000+2600+2400)(x0i^+y0j^)=[3000(0i^+0j^)+2600(0i^+0j^)+2400(32i^+10j^)]8000x0i^+8000y0j^=76800i^+24000j^

Equate i^ components of above expression.

x0=9.6 ft

Equate j^ components of the above expression.

y0=3 ft

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, −(60 mi/hr)i^ for v→C and (45 mi/hr)j^ for v→B in equation (4).

(3000+2600+2400)v→=[3000v→A+2600(45 mi/hr(1.467 ft/s1 mi/hr))j^+2400(−60 mi/hr(1.467 ft/s1 mi/hr))i^]8000v→=3000v→A+171639j^−211248i^v→=0.375v→A+21.455j^−26.406i^

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate j^ components of above expression.

59=(3+21.455t)56=21.455t

Rearrange the above expression for t.

t=2.61 s

The time elapsed from first collision to stop at pole is 2.61 s.

Conclusion:

Thus, the time elapsed from first collision to stop at pole is 2.61 s.

Expert Solution

To determine

(b)

Speed of car A.

Answer to Problem 14.20P

The velocity of car A is 73.3 mi/hr.

Explanation of Solution

Calculation:

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate the i^ components of both side in above expression.

46=[9.6+(0.375vA−26.406)t]36.4=(0.375vA−26.406)t

Substitute 2.61 s for t in above expression.

36.4=(0.375vA−26.406)(2.61)36.4=0.97875vA−68.92

Simplify above expression for vA.

vA=107.6 ft/s(0.681 mi/hr1 ft/s)=73.3 mi/hr

The velocity of car A is 73.3 mi/hr.

Conclusion:

Thus, the velocity of car A is 73.3 mi/hr.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

Time elapsed from first collision to the stop at P.

Answer to Problem 14.20P

The time elapsed from first collision to stop at pole is 2.61 s.

Explanation of Solution

Given:

Speed of car B is 45 mi/hr.

Speed of car C is 60 mi/hr.

Weight if car A is 3000 lb.

Weight of car B is 2600 lb.

Weight of car C is 2400 lb.

Coordinates of utility pole are (46 ft,59 ft).

Concept used:

Draw the diagram for the system mentioned as shown below:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.20P

Assume origin as point O where, Car A and Car B collides.

Write the expression for the mass center of first collision.

(mA+mB+mC)(x0i^+y0j^)=mA(xAi^+yAj^)+mB(xBi^+yBj^)+mC(xCi^+yCj^) ...... (1)

Multiply the whole equation (1) with g on both sides.

(WA+WB+WC)(x0i^+y0j^)=WA(xAi^+yAj^)+WB(xBi^+yBj^)+WC(xCi^+yCj^) ...... (2)

Here, WA is the weight of car A, WB is the weight of car B, WC is the weight of car C, (x0,y0) is coordinate of center of mass, (xA,yA) is the coordinate of car A, (xB,yB) is the coordinate of car B and (xC,yC) is the coordinate of car C.

Write the expression for conservation of momentum for the system.

(mA+mB+mC)v→=mAv→A+mBv→B+mCv→C ...... (3)

Multiply the equation (3) with g on both sides.

(WA+WB+WC)v→=WAv→A+WBv→B+WCv→C ...... (4)

Here, v→A is the velocity of car A, v→B is the velocity of car B, v→C is the velocity of car C and v→ is the velocity of center of mass of system.

Write the expression for the position of pole P.

xPi^+ypj^=x0i^+y0j^+v→t ...... (5)

Here, (xP,yP) is the coordinate of pole P.

Calculation:

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, 0 for xA, 0 for yA, 0 for xB, 0 for yB, 32 ft for xC and 10 ft for yC in equation (2).

(3000+2600+2400)(x0i^+y0j^)=[3000(0i^+0j^)+2600(0i^+0j^)+2400(32i^+10j^)]8000x0i^+8000y0j^=76800i^+24000j^

Equate i^ components of above expression.

x0=9.6 ft

Equate j^ components of the above expression.

y0=3 ft

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, −(60 mi/hr)i^ for v→C and (45 mi/hr)j^ for v→B in equation (4).

(3000+2600+2400)v→=[3000v→A+2600(45 mi/hr(1.467 ft/s1 mi/hr))j^+2400(−60 mi/hr(1.467 ft/s1 mi/hr))i^]8000v→=3000v→A+171639j^−211248i^v→=0.375v→A+21.455j^−26.406i^

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate j^ components of above expression.

59=(3+21.455t)56=21.455t

Rearrange the above expression for t.

t=2.61 s

The time elapsed from first collision to stop at pole is 2.61 s.

Conclusion:

Thus, the time elapsed from first collision to stop at pole is 2.61 s.

Expert Solution

To determine

(b)

Speed of car A.

Answer to Problem 14.20P

The velocity of car A is 73.3 mi/hr.

Explanation of Solution

Calculation:

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate the i^ components of both side in above expression.

46=[9.6+(0.375vA−26.406)t]36.4=(0.375vA−26.406)t

Substitute 2.61 s for t in above expression.

36.4=(0.375vA−26.406)(2.61)36.4=0.97875vA−68.92

Simplify above expression for vA.

vA=107.6 ft/s(0.681 mi/hr1 ft/s)=73.3 mi/hr

The velocity of car A is 73.3 mi/hr.

Conclusion:

Thus, the velocity of car A is 73.3 mi/hr.

Answer 8

Expert Solution

To determine

(a)

Time elapsed from first collision to the stop at P.

Answer to Problem 14.20P

The time elapsed from first collision to stop at pole is 2.61 s.

Explanation of Solution

Given:

Speed of car B is 45 mi/hr.

Speed of car C is 60 mi/hr.

Weight if car A is 3000 lb.

Weight of car B is 2600 lb.

Weight of car C is 2400 lb.

Coordinates of utility pole are (46 ft,59 ft).

Concept used:

Draw the diagram for the system mentioned as shown below:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.20P

Assume origin as point O where, Car A and Car B collides.

Write the expression for the mass center of first collision.

(mA+mB+mC)(x0i^+y0j^)=mA(xAi^+yAj^)+mB(xBi^+yBj^)+mC(xCi^+yCj^) ...... (1)

Multiply the whole equation (1) with g on both sides.

(WA+WB+WC)(x0i^+y0j^)=WA(xAi^+yAj^)+WB(xBi^+yBj^)+WC(xCi^+yCj^) ...... (2)

Here, WA is the weight of car A, WB is the weight of car B, WC is the weight of car C, (x0,y0) is coordinate of center of mass, (xA,yA) is the coordinate of car A, (xB,yB) is the coordinate of car B and (xC,yC) is the coordinate of car C.

Write the expression for conservation of momentum for the system.

(mA+mB+mC)v→=mAv→A+mBv→B+mCv→C ...... (3)

Multiply the equation (3) with g on both sides.

(WA+WB+WC)v→=WAv→A+WBv→B+WCv→C ...... (4)

Here, v→A is the velocity of car A, v→B is the velocity of car B, v→C is the velocity of car C and v→ is the velocity of center of mass of system.

Write the expression for the position of pole P.

xPi^+ypj^=x0i^+y0j^+v→t ...... (5)

Here, (xP,yP) is the coordinate of pole P.

Calculation:

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, 0 for xA, 0 for yA, 0 for xB, 0 for yB, 32 ft for xC and 10 ft for yC in equation (2).

(3000+2600+2400)(x0i^+y0j^)=[3000(0i^+0j^)+2600(0i^+0j^)+2400(32i^+10j^)]8000x0i^+8000y0j^=76800i^+24000j^

Equate i^ components of above expression.

x0=9.6 ft

Equate j^ components of the above expression.

y0=3 ft

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, −(60 mi/hr)i^ for v→C and (45 mi/hr)j^ for v→B in equation (4).

(3000+2600+2400)v→=[3000v→A+2600(45 mi/hr(1.467 ft/s1 mi/hr))j^+2400(−60 mi/hr(1.467 ft/s1 mi/hr))i^]8000v→=3000v→A+171639j^−211248i^v→=0.375v→A+21.455j^−26.406i^

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate j^ components of above expression.

59=(3+21.455t)56=21.455t

Rearrange the above expression for t.

t=2.61 s

The time elapsed from first collision to stop at pole is 2.61 s.

Conclusion:

Thus, the time elapsed from first collision to stop at pole is 2.61 s.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

Time elapsed from first collision to the stop at P.

Answer 13

Answer to Problem 14.20P

The time elapsed from first collision to stop at pole is 2.61 s.

Answer 14

Explanation of Solution

Given:

Speed of car B is 45 mi/hr.

Speed of car C is 60 mi/hr.

Weight if car A is 3000 lb.

Weight of car B is 2600 lb.

Weight of car C is 2400 lb.

Coordinates of utility pole are (46 ft,59 ft).

Concept used:

Draw the diagram for the system mentioned as shown below:

Vector Mechanics for Engineers: Dynamics, Chapter 14.1, Problem 14.20P

Assume origin as point O where, Car A and Car B collides.

Write the expression for the mass center of first collision.

(mA+mB+mC)(x0i^+y0j^)=mA(xAi^+yAj^)+mB(xBi^+yBj^)+mC(xCi^+yCj^) ...... (1)

Multiply the whole equation (1) with g on both sides.

(WA+WB+WC)(x0i^+y0j^)=WA(xAi^+yAj^)+WB(xBi^+yBj^)+WC(xCi^+yCj^) ...... (2)

Here, WA is the weight of car A, WB is the weight of car B, WC is the weight of car C, (x0,y0) is coordinate of center of mass, (xA,yA) is the coordinate of car A, (xB,yB) is the coordinate of car B and (xC,yC) is the coordinate of car C.

Write the expression for conservation of momentum for the system.

(mA+mB+mC)v→=mAv→A+mBv→B+mCv→C ...... (3)

Multiply the equation (3) with g on both sides.

(WA+WB+WC)v→=WAv→A+WBv→B+WCv→C ...... (4)

Here, v→A is the velocity of car A, v→B is the velocity of car B, v→C is the velocity of car C and v→ is the velocity of center of mass of system.

Write the expression for the position of pole P.

xPi^+ypj^=x0i^+y0j^+v→t ...... (5)

Here, (xP,yP) is the coordinate of pole P.

Calculation:

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, 0 for xA, 0 for yA, 0 for xB, 0 for yB, 32 ft for xC and 10 ft for yC in equation (2).

(3000+2600+2400)(x0i^+y0j^)=[3000(0i^+0j^)+2600(0i^+0j^)+2400(32i^+10j^)]8000x0i^+8000y0j^=76800i^+24000j^

Equate i^ components of above expression.

x0=9.6 ft

Equate j^ components of the above expression.

y0=3 ft

Substitute 3000 lb for WA, 2600 lb for WB, 2400 lb for WC, −(60 mi/hr)i^ for v→C and (45 mi/hr)j^ for v→B in equation (4).

(3000+2600+2400)v→=[3000v→A+2600(45 mi/hr(1.467 ft/s1 mi/hr))j^+2400(−60 mi/hr(1.467 ft/s1 mi/hr))i^]8000v→=3000v→A+171639j^−211248i^v→=0.375v→A+21.455j^−26.406i^

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate j^ components of above expression.

59=(3+21.455t)56=21.455t

Rearrange the above expression for t.

t=2.61 s

The time elapsed from first collision to stop at pole is 2.61 s.

Conclusion:

Thus, the time elapsed from first collision to stop at pole is 2.61 s.

Answer 15

Expert Solution

To determine

(b)

Speed of car A.

Answer to Problem 14.20P

The velocity of car A is 73.3 mi/hr.

Explanation of Solution

Calculation:

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate the i^ components of both side in above expression.

46=[9.6+(0.375vA−26.406)t]36.4=(0.375vA−26.406)t

Substitute 2.61 s for t in above expression.

36.4=(0.375vA−26.406)(2.61)36.4=0.97875vA−68.92

Simplify above expression for vA.

vA=107.6 ft/s(0.681 mi/hr1 ft/s)=73.3 mi/hr

The velocity of car A is 73.3 mi/hr.

Conclusion:

Thus, the velocity of car A is 73.3 mi/hr.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

Speed of car A.

Answer 20

Answer to Problem 14.20P

The velocity of car A is 73.3 mi/hr.

Answer 21

Explanation of Solution

Calculation:

Substitute 9.6 ft for x0, 3 ft for y0, 0.375v→A+21.455j^−26.406i^ for v→, 46 ft for xP, 59 ft for yP in equation (5).

46i^+59j^=9.6i^+3.0j^+(0.375v→A+21.455j^−26.406i^)t=9.6i^+3.0j^+0.375tv→A+21.455tj^−26.406ti^

Substitute vAi^ for v→A in above expression.

46i^+59j^=9.6i^+3.0j^+0.375tvAi^+21.455tj^−26.406ti^46i^+59j^=[9.6+(0.375vA−26.406)t]i^+(3+21.455t)j^

Equate the i^ components of both side in above expression.

46=[9.6+(0.375vA−26.406)t]36.4=(0.375vA−26.406)t

Substitute 2.61 s for t in above expression.

36.4=(0.375vA−26.406)(2.61)36.4=0.97875vA−68.92

Simplify above expression for vA.

vA=107.6 ft/s(0.681 mi/hr1 ft/s)=73.3 mi/hr

The velocity of car A is 73.3 mi/hr.

Conclusion:

Thus, the velocity of car A is 73.3 mi/hr.

Answer 22

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Knowing that the coordinates of the utility pole are x p = 46 ft and y p = 59 ft, determine ( a ) the time elapsed from the first collision to the stop at P , ( b ) the speed of car A .

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Answer to Problem 14.20P

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