3 m in v 1 v 2 55 m i h 70 m i h 1 h b when the faster

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Unformatted text preview: .0 m − 0 = = − 0.5 m s 4.0 s ∆t x 5 − x 1 0 − 4.0 m = = − 1.0 m s 4.0 s ∆t x5 − x0 0 − 0 = =0 5.0 s ∆t 22 x(m) 4 3 2 1 –1 –2 1 2 3 4 5 t(s) CHAPTER 2.8 2 ∆x . Thus, the difference in the times for the v two cars to complete the same 10 mile trip is ∆x ∆x ⎛ 10 m i 10 m i ⎞ ⎛ 60 m in ⎞ ∆t = t1 − t 2 = − =⎜ − ⎜ ⎟ = 2.3 m in v 1 v 2 ⎝ 55 m i h 70 m i h ⎟ ⎝ 1 h ⎠ ⎠ (a) The time for a car to make the trip is t = (b) When the faster car has a 15.0 min lead, it is ahead by a distance equal to that traveled by the slower car in a time of 15.0 min. This distance is given by ∆x 1 = v 2 ( ∆t ) = ( 55 m i h ) (15 m in ) . The faster car pulls ahead of the slower car at a rate of: v relat ive = 70 m i h − 55 m i h = 15 m i h . Thus, the time required for it to get distance ∆x 1 ahead is: ∆t = ∆x 1 v relative = ( 55 m i h ) (15 m in ) = 55 m in . 15.0 m i h Finally, the distance the faster car has traveled during this time is ⎛ 1 h ⎞ 64 m i ∆x = v t = ( 70 m i h ) ( 55 m in ) ⎜ = ⎝ 60 m in ⎟ ⎠ 2.9 (a) To correspond to only forward motion, the graph must always have a positive slope. Graphs of this type are a , e , a n d f . (b) For only backward motion, the slope of the graph must always be negative. Only graph c is of this type. (c) The slope of the graph must be constant in order to correspond to constant velocity. Graphs like this are d a n d f . (d) The graph corresponding to the largest constant velocity is the one with the largest constant slope. This is graph f . (e) To correspond to no motion (i.e., zero velocity), the slope of the graph must have a constant value of zero. This is graph d . 23 CHAPTER 2.10 2 The distance traveled by the space shuttle in one orbit is 2π ( Ea r t h ’s r a d iu s + 200 m iles ) = 2π ( 3963 + 200) m i = 2 .61 × 10 4 m i . Thus, the required time is 2.11 2.61 × 10 4 m i = 1.32 h . 19 800 m i h The total time for the trip is t = t1 + 22 .0 m in = t1 + 0.367 h , where t1 is the time spent...
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