COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 4, Problem 70P

(a)

To determine

The distance travelled by Carlos and Shannon when Shannon catches up to Carlos.

(a)

Expert Solution
Check Mark

Answer to Problem 70P

The distance travelled by Carlos and Shannon when Shannon catches up to Carlos is 77m_.

Explanation of Solution

The free body diagram of the given situation is shown in the Figure 1.

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 4, Problem 70P

Let the positive x direction be down the slope and the positive y direction is in the direction of normal force.

Since there is no force in the vertical direction, the net force in the vertical direction is zero.

    n=mgcosθ        (I)

Here, n is the normal force, m is the mass, g is the acceleration due to gravity, and θ is the angle makes the slope with horizontal.

From the free body diagram, write the expression for net force in the horizontal direction.

    mgsinθfk=max        (II)

Here, fk is the frictional force, and ax is the acceleration in the horizontal direction.

Use μkn for fk, equation (I) in (II), and solve for ax.

    mgsinθμkmgcosθ=maxax=g(sinθμkcosθ)        (III)

Write the kinematic equation to find the Carole’s velocity after travelling the 5m distance.

    vf,C2=vi,C2+2ax,CΔx        (IV)

Here, vf,C is the Carlos’s final velocity, vi,C is the Carlos’s initial velocity, ax,C is the Carlos’s initial acceleration, and Δx is the distance.

Write the kinematic equation to find the distance travelled by Carlos at the beginning pf the time interval.

    xf,C=xi,C+vi,Ct+12ax,Ct2        (V)

Here, xf,C is the final position of the Carlos, xi,C is the initial position of the Carlos, and t is the time taken by Shannon to reach Carlos.

Write the kinematic equation to find the distance travelled by Carlos at the beginning of the time interval.

    xf,S=xi,S+vi,St+12ax,St2        (VI)

Here, xf,S is the final position of the Shannon, and xi,S is the initial position of the Shannon.

Conclusion:

Substitute, 9.8m/s2 for g, 12° for θ, and 0.1 for μk in the equation (III) to find acceleration of Carlos.

    ax,C=9.8m/s2(sin12°0.1cos12°)=1.07894m/s2

Substitute, 9.8m/s2 for g, 12° for θ, and 0.01 for μk in the equation (III) to find acceleration of Shannon.

    ax,C=9.8m/s2(sin12°0.01cos12°)=1.94167m/s2

Substitute, 0 for vi,C, 5m for Δx, and 1.07894m/s2 for ax,C in the equation (IV), and solve for vf,C.

    vf,C2=0+2(1.07894m/s2)(5m)vf,C=3.28m/s

Substitute, 5m for xi,C, 3.28m/s for vi,C, and 1.07894m/s2 for ax,C in the equation (V), to find xf,C.

        xf,C=5m+(3.28m/s)t+12(1.07894m/s2)t2=5m+(3.28m/s)t+(0.53947m/s2)t2        (VII)

Substitute, 0 for xi,S, 0 for vi,S, and 1.94167m/s2 for ax,S in the equation (VI), to find xf,S.

        xf,S=0+0+12(84167m/s2)t2=0.970m/s2t2        (VIII)

The distance travelled by the Shannon and Carlos are same.

Equate equation (VII) and (VIII), and solve for t.

    5m+(3.28m/s)t+(0.53947m/s2)t2=0.970m/s2t2(0.4313m/s2)t2(3.28m/s)t5m=0t=8.9s

Substitute, 8.9s for t in the equation (VIII), to find xf,S.

    xf,S=0.970m/s2×(8.9s)2=77m

Therefore, the distance travelled by Carlos and Shannon when Shannon catches up to Carlos is 77m_.

(b)

To determine

The speed of the Shannon with respect to Carlos.

(b)

Expert Solution
Check Mark

Answer to Problem 70P

The speed of the Shannon with respect to Carlos is 4.4m/s_.

Explanation of Solution

Write the kinematic equation.

    v=u+at        (IX)

Conclusion:

Substitute, 0 for u, 1.94167m/s2 for a, and 8.9s for t in the equation (IX), to find the velocity of Shannon after time t (vf,S).

    vf,S=0+(1.94167m/s2)(8.9s)=17.28m/s

Substitute, 3.28m/s2 for u, 1.07894m/s2 for a, and 8.9s for t in the equation (IX), to find the velocity of Carlos after time t (vf,C).

    vf,C=3.28m/s2+(1.07894m/s2)(8.9s)=12.88m/s

Since Carlos is moving in the same direction as Shannon, the relative speed of Shannon with respect to Carlos can be calculated as follows,

    vSC=17.28m/s12.88m/s=4.4m/s

Therefore, the speed of the Shannon with respect to Carlos is 4.4m/s_.

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Chapter 4 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

Ch. 4.4 - Prob. 4.8PPCh. 4.5 - Prob. 4.5ACPCh. 4.5 - Prob. 4.10PPCh. 4.5 - Prob. 4.5BCPCh. 4.6 - Prob. 4.11PPCh. 4.6 - Prob. 4.6CPCh. 4 - Prob. 1CQCh. 4 - Prob. 2CQCh. 4 - Prob. 3CQCh. 4 - Prob. 4CQCh. 4 - Prob. 5CQCh. 4 - Prob. 6CQCh. 4 - Prob. 7CQCh. 4 - Prob. 8CQCh. 4 - Prob. 9CQCh. 4 - Prob. 10CQCh. 4 - Prob. 11CQCh. 4 - Prob. 12CQCh. 4 - Prob. 13CQCh. 4 - Prob. 14CQCh. 4 - Prob. 15CQCh. 4 - Prob. 16CQCh. 4 - Prob. 17CQCh. 4 - Prob. 18CQCh. 4 - Prob. 19CQCh. 4 - Prob. 1MCQCh. 4 - Prob. 2MCQCh. 4 - Prob. 3MCQCh. 4 - Prob. 4MCQCh. 4 - Prob. 5MCQCh. 4 - Prob. 6MCQCh. 4 - Prob. 7MCQCh. 4 - Prob. 8MCQCh. 4 - Prob. 9MCQCh. 4 - Prob. 10MCQCh. 4 - Prob. 11MCQCh. 4 - Prob. 12MCQCh. 4 - Prob. 13MCQCh. 4 - Prob. 14MCQCh. 4 - Prob. 15MCQCh. 4 - Prob. 16MCQCh. 4 - Prob. 17MCQCh. 4 - Prob. 18MCQCh. 4 - Prob. 19MCQCh. 4 - Prob. 20MCQCh. 4 - Prob. 21MCQCh. 4 - Prob. 1PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 61PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69PCh. 4 - Prob. 70PCh. 4 - Prob. 71PCh. 4 - Prob. 72PCh. 4 - Prob. 73PCh. 4 - Prob. 74PCh. 4 - Prob. 75PCh. 4 - Prob. 76PCh. 4 - Prob. 77PCh. 4 - Prob. 78PCh. 4 - Prob. 79PCh. 4 - Prob. 80PCh. 4 - Prob. 81PCh. 4 - Prob. 82PCh. 4 - Prob. 83PCh. 4 - Prob. 84PCh. 4 - Prob. 85PCh. 4 - Prob. 86PCh. 4 - Prob. 87PCh. 4 - Prob. 88PCh. 4 - Prob. 90PCh. 4 - Prob. 91PCh. 4 - Prob. 92PCh. 4 - Prob. 93PCh. 4 - Prob. 94PCh. 4 - Prob. 95PCh. 4 - Prob. 97PCh. 4 - Prob. 98PCh. 4 - Prob. 99PCh. 4 - Prob. 100PCh. 4 - Prob. 101PCh. 4 - Prob. 102PCh. 4 - Prob. 103PCh. 4 - Prob. 104PCh. 4 - Prob. 105P
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