1.2. POSITION. DISPLACEPJENT, VELOCITY 7 used throughout this book to denote a change in the quantity following the symbol. meaning the difference between its initial value and its nal value. The time interval itself will be written as AI and can be expressed as 31:1, l,- (1.2) where again i. and I; are the initial and nal values of the time parameter (imagine, for instance. that you are reading time in seconds on a digital clock, and you are interested in the change in the object's position between second 130 and second 132: then i, = 130 s' t; = 132 s, and At = 2 5]. You can practice reading oft displaoements [mm Figure 1.2. The displacement between .t' = 0.5s and if = Is. for instalme, is 0.25m (3:,- = 0.15m. 3:1 = 0.1m). ()n the other hand. between I, = 1s and I; = 1.3s. the displacement is An; = 07 0.1 = 70.111). Notice two important things about the displacement. First. it can be positive or negative Positive means the object moved. overall. in the positive direction; negative means it moved, overall. in the negative diiection. Second. even when it is positive. the displacement does not always equal the distance traveled by the object (distance. 01 courseI is always dened as a positive quantity]. because if the object \"doubles hack" on its tracks [or some distance, that distance does not count towards the overall displacement. as the following example shows: Example LB: For the object whose position is shown in Figure 1.2 (reproduced here): [a] \"that is the displacement between t, = ".5 s and if = 1.5 s? (b) \\Vhat is the distance traveled between :th t. =.5s and tf= 1.5s? in: Answer: (a) The position at the time t, = 0.5.9 is I.- : l'|.lm. and the position at the time 1! =153 is 1J- = 0.05m. By the deni- tion of displacement. then. A: marl.- : 0.05m(.15m} : mm (b) To calculate the total distance traveled, we need to look at the details of the motion in between those two times, and make sure we add (with a positive sign} every segment of the motion. regardless of whether it is in the positive or negative direction. As you can see from the gure, between the times (lists and 1.0s the object moves to the right (Le, in the positive 1 direction), from I : .lm to 1 : 0.1m. that is, a distance of [1.25 in. Then, between t = 1.0s and t = 1.55, the object moves to the left, from :c = (11m to :c = 0.l'|5m. This corresponds to a distance 0.15m. The total distance traveled is therefore :1 = 0.25 111+ .15 m= 0.4m. This is greater than the total displacement. because the object did "double back\" at I = 1.0 s. and so it only ended up 0.1m away From where it started. Multiple Choice Homework 2: Velocity. Reference Frames Problem 1: At which of these times is the instantaneous velocity of the object in Fig 1.2 zero? Select One of the Following (a) t i] 37:, [his and Lil; b) to (1.35. Is and l..5 ) t (1.5s and 1.25 (d) to (1.2 s and 1.35 c Problem 2: For the next Few questions. let E denote the Frame of reference of the Earth, A the "air" fame of reference, and P an airplane. Take the eastward direction to be positive. If the Wind around the plane is blowing due west at ~15 mph and the airplane is llying due east with an airspeed of Hull] mph, what 5 the value of in\"? [Using the convention for subscripts introduced in your textbook.) Select One of the Following (a) REM] rnph (b) :innmpi. (c) :15mph (d) 45rnph Problem 3: What is the value of [Sup-l Select One of the Following (a) Hill] rnph (b) 3l)[)mph (c) 4-3mph (d) ;l5mph Problem II: What is the value of tit-r4? Select One of the Following Problem 5: At some point the plane [which we are going to (all plane 1 now) crosses path with another plane [plane 2] that is traveling due west With an airspeed of 35" mph. What is the velocity of plane 2 relative to plane 1