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 4, Problem 75P

(a)

To determine

The drag racer time and the minimum coefficient of static friction between the tires and the road.

(a)

Expert Solution
Check Mark

Answer to Problem 75P

The drag racer time is 7.7s and the minimum coefficient of static friction between the tires and the road is 1.38.

Explanation of Solution

Write the expression for acceleration of the drag racer from Kinematic equation of motion.

    vf2=vi2+2as

Here, vf is the final velocity of the racer, vi is the initial velocity of the racer, a is the acceleration of the racer, and s is the distance of the finishing line of the race track.

Rewrite the above expression for acceleration.

vf2vi2=2asa=vf2vi22s        (I)

Write the expression for drag racer time from kinematic equation.

    vf=vi+at

Here, t is the drag racer time.

Rewrite the above expression for time.

vfvi=att=vfvia        (II)

Write the expression for frictional force.

f=μsmg        (III)

Here, f is the frictional force, μs is the minimum coefficient of static friction between the tires and road, m is the mass of the drag racer, and g is the acceleration due to gravity.

Write the expression for horizontal force exerted on the racer.

F=ma        (IV)

Here, F is the horizontal force and a is the acceleration of the racer.

Compare the above relation (III) and (IV) to find the coefficient of static friction.

μsmg=maμsg=aμs=ag        (V)

Conclusion:

Substitute 104m/s for vf, 400m for s, and 0m/s for vi in the equation (I) to find a.

a=(104m/s)2(0m/s)22(400m)=13.5m/s2

Substitute 104m/s for vf, 13.5m/s2 for a, and 0m/s for vi in the equation (II) to find t.

t=104m/s0m/s13.5m/s2=7.7s

Substitute 9.80m/s2 for g and 13.5m/s2 for a in the equation (V) to find μS.

μs=13.5m/s29.80m/s2=1.38

Thus, the drag racer time is 7.7s and the minimum coefficient of static friction between the tires and the road is 1.38.

(b)

To determine

The time taken to finish the race.

(b)

Expert Solution
Check Mark

Answer to Problem 75P

The time taken to finish the race is 7.76s.

Explanation of Solution

If the acceleration is reduced to 30% because of the bad tires or wet pavement, the new acceleration of the drag racer is,

a=30100a        (VI)

Here, a is the new acceleration of the racer.

The net acceleration of the racer is,

anet=aa        (VII)

Write the expression for the time taken to finish the race.

    t=vfanet        (VIII)

Here, t is the time taken to finish the race

Conclusion:

Substitute 13.5m/s2 for a in the equation (VI) to find a.

a=30100(13.5m/s2)=0.135m/s2

Substitute 0.135m/s2 for a and 13.5m/s2 for a in the equation (VII) to find anet.

anet=13.5m/s20.135m/s2=13.4m/s2

Substitute 104m/s for vf and 13.4m/s2 for anet in the equation (VIII) to find t.

t=104m/s13.4m/s2=7.76s

Thus, the time taken to finish the race is 7.76s.

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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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