Physics Solution Manual for 1100 and 2101

22 d the acceleration due to gravity points downward

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ngines shut down 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 positive. 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 PROBLEMS ______________________________________________________________________________ 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 displacement. 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 Distance 1500 m = = 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 1.5 m = = 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. ____________________________________...
View Full Document

Ask a homework question - tutors are online