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Unformatted text preview: 1 chapter 2 1. Assuming the horizontal velocity of the ball is constant, the horizontal displace- ment is Δ x = v Δ t where Δ x is the horizontal distance traveled, Δ t is the time, and v is the (horizontal) velocity. Converting v to meters per second, we have 160 km / h = 44 . 4 m / s. Thus Δ t = Δ x v = 18 . 4 m 44 . 4 m / s = 0 . 414 s . The velocity-unit conversion implemented above can be figured “from basics” (1000 m = 1 km, 3600 s = 1 h) or found in Appendix D. 2. Converting to SI units, we use Eq. 2-3 with d for distance. s avg = d t (110 . 6 km / h) 1000 m / km 3600 s / h ! = 200 . 0 m t which yields t = 6 . 510 s. We converted the speed km/h → m/s by converting each unit (km → m, h → s) individually. But we mention that the “one-step” conversion can be found in Appendix D (1 km / h = 0 . 2778 m / s). 3. We use Eq. 2-2 and Eq. 2-3. During a time t c when the velocity remains a positive constant, speed is equivalent to velocity, and distance is equivalent to displacement, with Δ x = v t c . (a) During the first part of the motion, the displacement is Δ x 1 = 40 km and the time interval is t 1 = (40 km) (30 km / h) = 1 . 33 h . During the second part the displacement is Δ x 2 = 40 km and the time interval is t 2 = (40 km) (60 km / h) = 0 . 67 h . Both displacements are in the same direction, so the total displacement is Δ x = Δ x 1 + Δ x 2 = 40 km + 40 km = 80 km. The total time for the trip is t = t 1 + t 2 = 2 . 00 h. Consequently, the average velocity is v avg = (80 km) (2 . 0 h) = 40 km / h . (b) In this example, the numerical result for the average speed is the same as the average velocity 40 km / h. (c) In the interest of saving space, we briefly describe the graph (with kilome- ters and hours understood): two contiguous line segments, the first having a slope of 30 and connecting the origin to ( t 1 , x 1 ) = (1 . 33 , 40) and the second having a slope of 60 and connecting ( t 1 , x 1 ) to ( t, x ) = (2 . 00 , 80). The average velocity, from the graphical point of view, is the slope of a line drawn from the origin to ( t, x ). 4. If the plane (with velocity v ) maintains its present course, and if the terrain continues its upward slope of 4 . 3 ◦ , then the plane will strike the ground after traveling Δ x = h tan θ = 35 m tan 4 . 3 ◦ = 465 . 5 m ≈ . 465 km . This corresponds to a time of flight found from Eq. 2-2 (with v = v avg since it is constant) t = Δ x v = . 465 km 1300 km / h = 0 . 000358 h ≈ 1 . 3 s . This, then, estimates the time available to the pilot to make his correction. 5. (a) Denoting the travel time and distance from San Antonio to Houston as T and D , respectively, the average speed is s avg 1 = D T = (55 km / h) T 2 + (90 km / h) T 2 T = 72 . 5 km / h which should be rounded to 73 km/h....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
- Fall '08