1.2 UNIFORM MOTION PRACTICE

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1 Making Connections 6. car s odometer measures the distance travelled. Its speedometer measures the instantaneous velocity. 7. Sign (d) is the best because it indicates the maximum allowed velocity (as opposed to a possible interpretation of velocity as average velocity). Second, the unit of km/h is proper SI form. unit kph is not recognized and the velocity of 60 omits units completely..2 UNIFORM MOTION (Pages 2 4). An object that falls straight down experiences linear motion but not uniform motion because it is accelerating as it falls. 2. d 2. m [S] d d 2 d d m [S] d 9.7 m [S] 2. m [S]? d 7.6 m [S] curling rock s displacement is 7.6 m [S]. 3. d 2.8 m [W] d d 2 d d m [E] d 2.6 m [E] 2.8 m [W]? d 5.4 m [E] dog s required displacement is 5.4 m [E]. 4. (a) d 0.0 m d d 2 d d m [fwd] d 4.4 m [fwd] 0.0 m? d 4.4 m [fwd] (b) d 4.4 m [fwd] d d 2 d d m [fwd] d 8.8 m [fwd] 4.4 m [fwd]? d 4.4 m [fwd] (c) d 4.4 m [fwd] d d 2 d d m [fwd] d 3.2 m [fwd] 4.4 m [fwd]? d 8.8 m [fwd] 5. y are the same. 6. d 50.0 m [fwd] v 6.9 s av d v av? 50.0 m [fwd] 6.9s v av 2.96 m/s [fwd] athlete s average velocity was 2.96 m/s [fwd]. 7. d v av (), vav 8. v av 2.4 mm/s [fwd] d v av () 40 s d 2.4 mm/s [fwd](40 s)? mm [fwd] d 34 cm [fwd] snail s displacement was 34 cm [fwd]. 9. v av 20.8 m/s [fwd] (assumed) d 78 m [fwd]? t t It would take the record holder 8.56 s. vav 78 m [fwd] 20.8 m/s [fwd] 8.56 s 6 Unit Forces and Motion Copyright 2002 Nelson Thomson Learning

2 Try This Activity: Attempting Uniform Motion (Page 4) (a) If the motion is uniform, the position-time graph is a straight line with a positive or negative slope. ( corresponding velocity-time graph is a straight horizontal line.) (b) It is fairly easy to use a battery-powered toy vehicle or a glider on an air table to produce uniform velocity; however, in both cases, the moving object should be on a level surface to prevent slowing down or speeding up. It is much more difficult for a person to create uniform motion; in fact, it is impossible when starting or stopping. Walking backwards is also a problem. (Pages 5 6) Understanding Concepts 0. d 08 m [W] 72 m [W] d 36 m [W] 2.0 s m? m d 36m [W] 2.0s m 8 m/s [W] slope from 2.0 s to 4.0 s is the same as the slope of the entire line.. (a) Position (m [S]) Time (s) (b) (c) slope: m t m [S] 2.0s m m/s [S] All line segments have the same slope, indicating that the velocity is constant. Copyright 2002 Nelson Thomson Learning Chapter Motion 7

3 (d) (e) area length width m/s [S] (2.0 s) area. 0 4 m [S] This area represents the jet s displacement during the time interval. 2. (a) m d 5.0 m [E] 0.0s m m/s [E] (b) m d 5.0 m [E] 0.0s m m/s [E] (c) m d 0.0 m [W] 0.20s m m/s [W] 3. (a) area under graph 40.0 m/s [N](3.0 s) m [N] (b) area under graph 30.0 m/s [N](4.0 s) m [N] (c) area under graph 5.0 m/s [N](8.0 s) m [N] m [S] Section.2 Questions (Pages 7 8). (a) slope of a line on a position time graph represents average velocity. (b) area under a line on a velocity time graph represents displacement. 2. magnitude of the slope of the line on a position time graph and the magnitude of the velocity of the motion are equivalent. 3. (a) runner s average velocity should be greater in the 60.0-m sprint than in the 50.0-m sprint since the athlete is able to maintain the top velocity for a greater distance (and time). This assumes the period of acceleration from the start of the race to the top velocity is the same for both sprints. 8 Unit Forces and Motion Copyright 2002 Nelson Thomson Learning

4 (b) 50.0-m sprint: d 50.0 m [E] v av d 5.96 s v av? 50.0 m 5.96s v av 8.39 m/s [E] 60.0-m sprint: d 60.0 m [E] v av d 6.92 s v av? 60.0 m 6.92s v av 8.67 m/s [E] runner has a greater average velocity when competing in the 60.0 m sprint. 4. v av 3.30 km/h [W] d d km [W] v av? km[w] 3.30 km / h[w] h 0 d, 0.2 h Kon-Tiki expedition took 0 d, 0.2 h to complete m [fwd] 6. v av 8.50 m/s [fwd]? d v av 0.0m [fwd] 8.50m / s [fwd] 2.9 s hurdle race would take 2.9 s to complete. v av 90.0 km/h [E] 25 m/s [E] v av ( ).0 s (the time interval between the first and second riders) 25 m/s [E](.0 s)? 25 m [E] (or km [E]) second rider has a displacement of 25 m [W] of the first rider (and slightly south as well, being in the other lane). Applying Inquiry Skills 7. (a) Puck A has the higher average velocity because it travels farther than puck B during the same time interval. (b) Time (s) d A (m[fwd]) d B (m[fwd]) (c) Puck A appears to travel with uniform motion. All points fall along a straight line on the position time graph. Puck B appears not to travel with uniform motion. When a line of best fit is drawn, not all points fall on the line. This indicates that puck B does not always travel equal distances in equal time intervals. puck may have encountered wrinkles in the paper that could have affected the motion. Copyright 2002 Nelson Thomson Learning Chapter Motion 9

5 (d) average velocity slope of position time graph. slope d 7.50 cm [fwd] 0.60s slope 2 cm/s [fwd] (or 2.5 cm/s [fwd] if 3 significant digits are carried) (e) area under graph: A l w 0.60 s(2.5 cm/s [fwd]) A 7.5 cm [fwd] This area represents the displacement of puck A during the recorded time interval. (f) sources of error: - imprecise measurements of distances using the ruler - unknown precision of the period of the spark timer - imprecision involved with the plotting of data on the position time graph - factors that could have influenced the motion of the pucks include: - contact with the paper on the air table - motion of the pucks restricted by the air hoses - ambient air currents affecting pucks.3 TWO-DIMENSIONAL MOTION (Pages 9 23) 2. If the vectors are added in any order, the resultant is the same. For example, if the vectors are added in the reverse order as shown below: 8.0 km [W] km [S] km [E] 3. (b) d 26 m [N] d 2 36 m [E] d R? ( R ) 2 ( ) 2 + ( 2 ) 2 R ( ) 2 + ( 2 ) 2 (26 m) 2 + (36 m) 2 R 44 m resultant displacement is 44 m [36 N of E] θ tan 2 36 m tan 26 m θ 54 d R 0 Unit Forces and Motion Copyright 2002 Nelson Thomson Learning