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does the free-fall motion begin. ( ) 20. (c) According to an equation of kinematics v = v0 + 2ax, with v = 0 m/s , the launch
2 2 speed v0 is proportional to the square root of the maximum height.
21. (a) An equation of kinematics ( v = v0 + at ) gives the answer directly.
22. (d) The acceleration due to gravity points downward, in the same direction as the initial
velocity of the stone thrown from the top of the cliff. Therefore, this stone picks up speed as
it approaches the nest. In contrast, the acceleration due to gravity points opposite to the
initial velocity of the stone thrown from the ground, so that this stone loses speed as it
approaches the nest. The result is that, on average, the stone thrown from the top of the cliff
travels faster than the stone thrown from the ground and hits the nest first.
23. 1.13 s 44 KINEMATICS IN ONE DIMENSION 24. (a) The slope of the line in a position versus time graph gives the velocity of the motion.
The slope for part A is positive. For part B the slope is negative. For part C the slope is
25. (b) The slope of the line in a position versus time graph gives the velocity of the motion.
Section A has the smallest slope and section B the largest slope.
26. (c) The slope of the line in a position versus time graph gives the velocity of the motion.
Here the slope is positive at all times, but it decreases as time increases from left to right in
the graph. This means that the positive velocity is decreasing as time increases, which is a
condition of deceleration. Chapter 2 Problems 45 CHAPTER 2 KINEMATICS IN ONE DIMENSION
1. SSM REASONING AND SOLUTION
a. The total distance traveled is found by adding the distances traveled during each segment
of the trip.
6.9 km + 1.8 km + 3.7 km = 12.4 km b. All three segments of the trip lie along the east-west line. Taking east as the positive
direction, the individual displacements can then be added to yield the resultant
6.9 km + (–1.8 km) + 3.7 km = +8.8 km
The displacement is positive, indicating that it points due east. Therefore,
Displacement of the whale = 8.8 km, due east
2. REASONING The displacement is a vector that points from an object’s initial position to
its final position. If the final position is greater than the initial position, the displacement is
positive. On the other hand, if the final position is less than the initial position, the
displacement is negative. (a) The final position is greater than the initial position, so the
displacement will be positive. (b) The final position is less than the initial position, so the
displacement will be negative. (c) The final position is greater than the initial position, so
the displacement will be positive.
SOLUTION The displacement is defined as Displacement = x – x0, where x is the final
position and x0 is the initial position. The displacements for the three cases are:
(a) Displacement = 6.0 m − 2.0 m = +4.0 m
(b) Displacement = 2.0 m − 6.0 m = −4.0 m
(c) Displacement = 7.0 m − (−3.0 m) = +10.0 m 3. SSM REASONING The average speed is the distance traveled divided by the elapsed
time (Equation 2.1). Since the average speed and distance are known, we can use this
relation to find the time. 46 KINEMATICS IN ONE DIMENSION SOLUTION The time it takes for the continents to drift apart by 1500 m is
= 5 × 104 yr
Average speed cm 1 m 3 yr 100 cm ______________________________________________________________________________
Elapsed time = 4. REASONING Average speed is the ratio of distance to elapsed time (Equation 2.1), so the
elapsed time is distance divided by average speed. Both the average speed and the distance
are given in SI base units, so the elapsed time will come out in seconds, which can then be
converted to minutes (1 min = 60 s).
SOLUTION First, calculate the elapsed time ∆t in seconds: ∆t = Distance
= 140 s
Average speed 1.1×10−2 m/s (2.1) Converting the elapsed time from seconds to minutes, we find that 1 min ∆t = 140 s = 2.3 min 60 s ( 5. ) x − x0 REASONING According to Equation 2.2 v = , the average velocity ( v ) is equal t − t0 to the displacement ( x − x0 ) divided by the elapsed time ( t − t0 ) , and the direction of the
average velocity is the same as that of the displacement. The displacement is equal to the
difference between the final and initial positions.
SOLUTION Equation 2.2 gives the average velocity as v= x − x0
t − t0 Therefore, the average velocities for the three cases are:
(a) Average velocity = (6.0 m − 2.0 m)/(0.50 s) = +8.0 m/s
(b) Average velocity = (2.0 m − 6.0 m)/(0.50 s) = −8.0 m/s
(c) Average velocity = [7.0 m − (−3.0 m)]/(0.50 s) = +20.0 m/s Chapter 2 Problems 47 The algebraic sign of the answer conveys the direction in each case.
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