51_PartUniversity Physics Solution

51_PartUniversity Physics Solution - Motion Along a...

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Motion Along a Straight Line 2-21 (d) area h = under v - t graph. () 3 Triangle 1 (0.5 ms) (0.5 ms) 33 cm/s 8.3 10 cm 2 hA ≈= = × . 2 Triangle 1 (1.0 ms) (1.0 ms)(100 cm s) 5.0 10 cm 2 = × . Triangle Rectangle 1 (1.5 ms) (1.3 ms) 133 cm/s (0.2 ms)(1.33) 0.11 cm 2 A ≈+ = = EVALUATE: The acceleration is constant until 1.3 ms t = , and then it is zero. 2 980 cm/s g = . The acceleration during the first 1.3 ms is much larger than this and gravity can be neglected for the portion of the jump that we are considering. Figure 2.52 2.53. (a) IDENTIFY and SET UP: The change in speed is the area under the x a versus t curve between vertical lines at 2.5 s t = and 7.5 s. t = EXECUTE: This area is 22 1 2 (4.00 cm/s 8.00 cm/s )(7.5 s 2.5 s) 30.0 cm/s +− = This acceleration is positive so the change in velocity is positive. (b) Slope of x v versus t is positive and increasing with t . The graph is sketched in Figure 2.53. Figure 2.53 EVALUATE: The calculation in part (a) is equivalent to av- . xx va t Δ Since x a is linear in t , av- 0 / 2 . x aa a =+ Thus 1 av- 2 (4.00 cm/s 8.00 cm/s ) x a for the time interval 2.5 s t = to 7.5 s. t = 2.54. IDENTIFY: The average speed is the total distance traveled divided by the total time. The elapsed time is the distance traveled divided by the average speed. SET UP: The total distance traveled is 20 mi. With an average speed of 8 mi/h for 10 mi, the time for that first 10 miles is 10 mi 1.25 h 8 mi/h = . EXECUTE: (a) An average speed of 4 mi/h for 20 mi gives a total time of 20 mi 5.0 h 4 mi/h = . The second 10 mi must be covered in 5.0 h 1.25 h 3.75 h −= . This corresponds to an average speed of 10 mi 2.7 mi/h 3.75 h = . (b) An average speed of 12 mi/h for 20 mi gives a total time of 20 mi 1.67 h 12 mi/h = . The second 10 mi must be covered in 1.67 h 1.25 h 0.42 h . This corresponds to an average speed of 10 mi 24 mi/h 0.42 h = . (c) An average speed of 16 mi/h for 20 mi gives a total time of 20 mi 1.25 h 16 mi/h = . But 1.25 h was already spent during the first 10 miles and the second 10 miles would have to be covered in zero time. This is not possible and an average speed of 16 mi/h for the 20-mile ride is not possible. EVALUATE: The average speed for the total trip is not the average of the average speeds for each 10-mile segment. The rider spends a different amount of time traveling at each of the two average speeds. 2.55. IDENTIFY: x dx vt dt = and x x dv a dt = . SET UP: 1 nn d tn t dt = , for 1 n . EXECUTE: (a) 32 2 ( ) (9.00 m/s ) (20.0 m/s ) 9.00 m/s x t t =− + . ( ) (18.0 m/s ) 20.0 m/s x at t . The graphs are sketched in Figure 2.55.
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2-22 Chapter 2 (b) The particle is instantaneously at rest when ( ) 0 x vt = . 0 0 x v = and the quadratic formula gives 2 1 (20.0 (20.0) 4(9.00)(9.00)) s 1.11 s 0.48 s 18.0 t = ± . 0.63 s t = and 1.59 s t = . These results agree with the - x graphs in part (a). (c) For 0.63 s t = , 32 2 (18.0 m/s )(0.63 s) 20.0 m/s 8.7 m/s x a =− = . For 1.59 s t = , 2 8.6 m/s x a =+ . At 0.63 s t = the slope of the - x graph is negative and at 1.59 s t = it is positive, so the same answer is deduced from the () x graph as from the expression for ( ) x at .
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51_PartUniversity Physics Solution - Motion Along a...

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