COLLEGE PHYSICS,VOL.1
COLLEGE PHYSICS,VOL.1
2nd Edition
ISBN: 9781111570958
Author: Giordano
Publisher: CENGAGE L
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Chapter 3, Problem 83P

(a)

To determine

The reaction time.

(a)

Expert Solution
Check Mark

Answer to Problem 83P

The reaction time is 0.30 s

Explanation of Solution

Write the equation of motion.

  yy0=v02+12at2        (I)

Here, v0 is the final velocity, t is the time, a is the acceleration, and  yy0 is the fallen distance.

Conclusion:

Substitute 0 for v0, g for a,in expression (I)

    yy0 =12gt2        (II)

Substitute 0.45m for yy0, 9.8m/s2 for g,in expression (II) and solve for time,

  t=2(yy0)g=2(0.45m)(9.8m/s2)=0.30 s

The reaction time is 0.30 s

(b)

To determine

The distance for each time intervals.

(b)

Expert Solution
Check Mark

Answer to Problem 83P

The distance for each time intervals are, for t=0.14 s, 10 cm, for t=0.16 s, 13 cm, for t=0.18 s, 16 cm, for t=0.20 s, 20 cm and for t=0.22 s, 24 cm.

Explanation of Solution

Write the equation of motion.

  yy0=v02+12at2        (I)

Here, v0 is the final velocity, t is the time, a is the acceleration, and  yy0 is the fallen distance.

Conclusion:

Substitute 0 for v0, g for a,in expression (I)

    yy0 =12gt2y0y=4.9×t2        (II)

Substitute 0.14sec for t, in expression (II) and solve for distance,

    y0y=4.9×(0.14sec)2=10cm

Substitute 0.16sec for t, in expression (II) and solve for distance,

    y0y=4.9×(0.16sec)2=13cm

Substitute 0.18sec for t, in expression (II) and solve for distance,

    y0y=4.9×(0.18sec)2=16cm

Substitute 0.20sec for t, in expression (II) and solve for distance,

    y0y=4.9×(0.20sec)2=20cm

Substitute 0.22sec for t, in expression (II) and solve for distance,

    y0y=4.9×(0.22sec)2=24cm

The distance for each time intervals are, for t=0.14 s, 10 cm, for t=0.16 s, 13 cm, for t=0.18 s, 16 cm, for t=0.20 s, 20 cm and for t=0.22 s, 24 cm.

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Chapter 3 Solutions

COLLEGE PHYSICS,VOL.1

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