Physics 16 Problem Set 2 Solutions

!"#$%&$'()'!*+,-./'0.1'2'0+-31%+4$' !5*1'('+6'7*%11.4' 2.15. IDENTIFY and SET UP: Use and to calculate and EXECUTE: (a) At (b) Set and solve for t: (c) Set and solve for t. This gives point after 32.0 s. (d) Turtle is 10.0 cm from starting point when Set and solve for t: and At and The turtle returns to the starting or At Set and solve for t: (other root to the quadratic equation is negative and hence nonphysical). At (e) The graphs are sketched in Figure 2.15. Figure 2.15 EVALUATE: The acceleration is constant and negative. is linear in time. It is initially positive, decreases to zero, and then becomes negative with increasing magnitude. The turtle initially moves farther away from the origin but then stops and moves in the 2.34. IDENTIFY: Use constant acceleration equations to find SET UP: Let be the direction the train is traveling. EXECUTE: At for each segment of the motion. to 14.0 s: . , the speed is . In the next 70.0 s, and . For the interval during which the train is slowing down, gives , and . The total distance traveled is . EVALUATE: The acceleration is not constant for the entire motion but it does consist of constant acceleration segments and we can use constant acceleration equations for each segment. ! . 2.44. IDENTIFY: Apply constant acceleration equations to the vertical motion of the sandbag. SET UP: Take upward. . The initial velocity of the sandbag equals the velocity of the balloon, so . When the balloon reaches the ground, maximum height the sandbag has EXECUTE: (a) . At its . : . The sandbag is 40.9 m above the ground. . : . The sandbag is 40.1 m above the ground. . (b) , , . gives . and . t must be positive, so . (c) (d) , , . gives . The maximum height is 41.3 m above the ground. (e) The graphs of , , and y versus t are given in Fig. 2.44. Take at the ground . EVALUATE: The sandbag initially travels upward with decreasing velocity and then moves downward with increasing speed. Figure 2.44 ! 2.95. IDENTIFY: Apply constant acceleration equations to the motion of the two objects, the student and the bus. SET UP: For convenience, let the student's (constant) speed be and the bus's initial position be Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero, and the initial velocity of the bus is taken to be zero. The positions of the student and the bus as functions of time are then and EXECUTE: (a) Setting and solving for the times t gives . . The student will be likely to hop on the bus the first time she passes it (see part (d) for a discussion of the later time). During this time, the student has run a distance (b) The speed of the bus is . (c) The results can be verified by noting that the x lines for the student and the bus intersect at two points, as shown in Figure 2.95a. (d) At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then catches up to her. At this later time the bus's velocity is (e) No; , and the roots of the quadratic are imaginary. When the student runs at Figure 2.95b shows that the two lines do not intersect: (f) For the student to catch the bus, and so the minimum speed is She would be running for a time and covers a distance However, when the student runs at the lines intersect at one point, at , as shown in Figure 2.95c. EVALUATE: The graph in part (c) shows that the student is traveling faster than the bus the first time they meet but at the second time they meet the bus is traveling faster. Figure 2.95 3.4. IDENTIFY: . This vector will make a components are equal. SET UP: . EXECUTE: EVALUATE: -angle with both axes when its x- and y- . gives Both components of . change with t. ! 3.13. IDENTIFY: The car moves in projectile motion. The car travels during the time it travels 61.0 m horizontally. SET UP: Take to be downward. , . EXECUTE: downward , . Use the vertical motion to find the time in the air: gives Then (b) EVALUATE: gives since . . . . We calculate the final velocity by calculating its x and y components. !5*1'2'+6'7*%11.4' (a) The graph of acceleration vs time looks like: Integrating once gives: (notice Iʼve changed the units from m/s to mph) Integrating again gives: (b) According to this data, I got up to about 53 mph; by my speedometer it was a little over 50 mph. Note that the final speed is, according to the integral, about 0.2 mph, while I told you that I braked to a stop. This makes me think that the accuracy of the speed data is not much better than +/- 0.2 mph (0.1 m/s) by the time we've integrated for a while. (c) According to this data, I went about 170 m.!