Physics 107 HOMEWORK ASSIGNMENT #1
Cutnell & Johnson, 7th edition
Chapter 2: Conceptual Question 12
Chapter 2: Problems 16, 28, 30, 32
12 A ball is dropped from rest from the top of a building and strikes the ground with a speed vf. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is v0=vf, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building, above the halfway point, or below the halfway point. Give your reasoning.
16 An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1, the magnitude of the car’s acceleration is 3.0 m/s2. The magnitude of the car’s velocity at the end of stage 2 is 2.5 times greater than it is at the end of stage 1. Find the magnitude of the acceleration in stage 2.
*28 Review Interactive LearningWare 2.2 in preparation for this problem. A race driver has made a pit stop to refuel. After refueling, he leaves the pit area with an acceleration whose magnitude is 6.0 m/s2; after 4.0 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant speed of 70.0 m/s overtakes and passes the entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?
*30 In a historical movie, two knights on horseback start from rest 88.0 m apart and ride directly toward each other to do battle. Sir George’s acceleration has a magnitude of 0.300 m/s2, while Sir Alfred’s has a magnitude of 0.200 m/s2. Relative to Sir George’s starting point, where do the knights collide?
*32 Multiple-Concept Example 9 illustrates the concepts that are pertinent to this problem. A cab driver picks up a customer and delivers her 2.00 km away, on a straight route. The driver accelerates to the speed limit and, on reaching it, begins to decelerate at once. The magnitude of the deceleration is three times the magnitude of the acceleration. Find the lengths of the acceleration and deceleration phases.
SOLUTIONS:
12. REASONING AND SOLUTION The first ball has an initial speed of zero, since it is dropped from rest. It picks up speed on the way down, striking the ground at a speed . The second ball has a motion that is the reverse of that of the first ball. The second ball starts out with a speed and loses speed on the way up. By symmetry, the second ball will come to a halt at the top of the building. Thus, in approaching the crossing point, the second ball travels faster than the first ball. Correspondingly, the second ball must travel farther on its way to the crossing point than the first ball does. Thus, the crossing point must be located in the upper half of the building.
16. REASONING AND SOLUTION The velocity of the automobile for each stage is given by Equation 2.4: . Therefore,
Since the magnitude of the car's velocity at the end of stage 2 is 2.5 times greater than it is at the end of stage 1, . Thus, rearranging the result for v2, we find
28. REASONING AND SOLUTION The car enters the speedway with a speed of
v01 = a1t1 = (6.0 m/s2)(4.0 s) = 24 m/s
After an additional time, t, it will have traveled a distance of
x = v01t + a1t2/2
to overtake the other car. This second car travels the same distance x = v02t.
Equating and solving for t yields
30. REASONING The drawing shows the two knights, initially separated by the displacement d, traveling toward each other. At any moment, Sir George’s displacement is xG and that of Sir Alfred is xA. When they meet, their displacements are the same, so xG = xA.
According to Equation 2.8, Sir George's displacement as a function of time is
(1)
where we have used the fact that Sir George starts from rest
Since Sir Alfred starts from rest at x = d at t = 0 s, we can write his displacement as (again, employing Equation 2.8)
(2)
Solving Equation 1 for t2 and substituting this expression into Equation 2 yields
(3)
Noting that xA = xG when the two riders collide, we see that Equation 3 becomes
Solving this equation for xG gives .
SOLUTION Sir George’s acceleration is positive since he starts from rest and moves to the right (the positive direction). Sir Alfred’s acceleration is negative since he starts from rest and moves to the left (the negative direction). The displacement of Sir George is, then,
32. REASONING AND SOLUTION The distance covered by the cab driver during the two phases of the trip must satisfy the relation
(1)
where x1 and x2 are the displacements of the acceleration and deceleration phases of the trip, respectively. The quantities x1 and x2 can be determined from Equation 2.9 :
and
with and . Thus,
so that
(2)
Combining (1) and (2), we have,
Therefore, , and from Equation (1), . Thus, the length of the acceleration phase of the trip is , while the length of the deceleration phase is .