PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
10th Edition
ISBN: 9781337888462
Author: SERWAY
Publisher: CENGAGE L
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Chapter 4, Problem 37AP

Lisa in her Lamborghini accelerates at ( 3.00 i ^ 2.00 j ^ ) m / s 2 , while Jill in her Jaguar accelerates at ( 1.00 i ^ 3.00 j ^ ) m / s 2 . They both start from rest at the origin. After 5.00 s, (a) what is Lisa’s speed with respect to Jill. (b) how far apart are they, and (c) what is Lisa’s acceleration relative to Jill?

(a)

Expert Solution
Check Mark
To determine

The relative speed of Lisa with respect to Jill after 5.00sec.

Answer to Problem 37AP

The relative speed of Lisa with respect to Jill after 5.00sec is 26.9m/s.

Explanation of Solution

The acceleration of Lisa in her Lamborghini is (3.00i^2.00j^)m/s2 and acceleration of Jill is (1.00i^+3.00j^)m/s2.

Write the expression for the first law of motion to calculate the final velocity of Lisa’s Lamborghini

    vL=uL+aLt

Here, uL is the initial velocity of, t is the time, vL is the Lisa’s final velocity of and aL is the Lisa’s acceleration.

Substitute 0m/s for uL, 5.00sec for t  and (3.00i2.00j^)m/s2 for aL to find vL.

    vL=0m/s+(3.00i2.00j^)m/s2×5.00sec=(15.0i10.0j^)m/s

Write the expression for the first law of motion to calculate the final velocity of Jill’s Jaguar

    vJ=uJ+aJt

Here, uJ is the initial velocity of Jill, vJ is the final velocity of Jill and aJ is the acceleration of Jill.

Substitute 0m/s for uJ, 5.00sec for t and (1.00i+3.00j^)m/s2 for aJ to find the vJ.

    vJ=0m/s+(1.00i+3.00j^)m/s2×5.00sec=(5.0i+15.0j^)m/s

The relative velocity of Lisa with respect to Jill

    vLJ=vLvJ

Here, vLJ is the relative velocity of Lisa with respect to Jill.

Conclusion:

Substitute (15.0i10.0j^)m/s for vL and (5.00i+15.0j^)m/s for vJ to find vLJ.

  vLJ=(15.0i10.0j^)m/s(5.00i+15.0j^)m/s=(10.0i25.0j^)m/s

Find the magnitude of the above found velocity

  |vLJ|=10.02+(25.0)2=26.9m/s

Therefore, the Lisa’s speed with respect to Jill after 5.00sec is 26.9m/s.

(b)

Expert Solution
Check Mark
To determine

The distance between Lisa and Jill.

Answer to Problem 37AP

The distance between Lisa and Jill is 67.3m.

Explanation of Solution

Write the formula for second law of motion to calculate the displacement of Lisa’s Lamborghini

    sL=uLt+12aLt2

Here, sL is the position of Lisa’s.

Substitute 0m/s for uL 5.00sec for t and (3.00i2.00j^m/s2)m/s2 for aL to find the sL.

    sL=0m/st+12(3.00i2.00j^)m/s2(5.00sec)2=(37.5i25j^)m

Write the formula for second law of motion to calculate the displacement of Jill’s Jaguar

    sJ=uJt+12aJt2

Here, sJ is the position of Jill’s Jaguar.

Substitute 0m/s for uJ 5.00sec for t and (1.00i+3.00j^)m/s2 for aJ to find sJ.

    sJ=0m/st+12(1.00i+3.00j^)m/s2(5.00sec)2=(12.5i37.5j^)m

Write the expression for the separation between these persons

    sLJ=sLsJ

Here, sLJ is the relative position of Lisa with respect to Jill.

Substitute (37.5i25j^)m for sL and (12.5i37.5j^)m for sJ to find sLJ.

    sLJ=(37.5i25j^)m(12.5i37.5j^)m=(25i62.5j^)m|sLJ|=252+(62.5)2=67.3m

Conclusion:

Therefore, the distance between Lisa and Jill is 67.3m.

(c)

Expert Solution
Check Mark
To determine

The relative acceleration of Lisa with respect to Jill.

Answer to Problem 37AP

The relative acceleration of Lisa with respect to Jill is (2.00i5.00j^)m/s2.

Explanation of Solution

Write the formula to calculate Lisa acceleration relative to Jill

    aLJ=aLaJ

Here, aLJ is the relative acceleration of Lisa with respect to Jill.

Conclusion:

Substitute (3.00i2.00j^)m/s2 for aL and (1.00i+3.00j^)m/s2 for aJ to find aLJ.

    aLJ=(3.00i2.00j^)m/s2(1.00i+3.00j^)m/s2=(2.00i5.00j^)m/s2

Therefore, Lisa’s acceleration relative to Jill is (2.00i5.00j^)m/s2

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Chapter 4 Solutions

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN

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