Assumed the acceleration was constant and that the receiver could be modeled as a point particle.

Transcription

1 PUM Physics II - Kinematics Lesson 16 Solutions Page 1 of Regular Problem v o = 10 m/s v = -2.0 m/s t = s v = v o + at -2.0 m/s = (10 m/s) + a(0.020 s) a = (-12 m/s)/(0.020 s) = -600 m/s 2 Assumed the acceleration was constant and that the receiver could be modeled as a point particle Regular Problem Assuming you travel straight, hit the brakes exactly 0.80 s when traffic is 30 m in front and acceleration is constant. v o = 24 m/s a = -8.0 m/s 2 How far do you travel during the 0.80 s needed to react: x(t) = (24 m/s)t x = (24 m/s)(0.80 s) = 19 m. How long does it take to come to a complete stop after brakes are pressed: v = v o + at 0 m/s = (24 m/s) + (-8.0 m/s 2 )t t = (-24 m/s)/(-8.0 m/s 2 ) = 3.0 s How far do you travel during the 3 s needed to stop: x(t) = (24 m/s)t + ½(-8.0 m/s 2 )t 2 x = (24 m/s)(3.0 s) + ½(-8.0 m/s 2 )(3.0 s) 2 = 36 m You need 55 m to stop. You better call the insurance company 16.3 Represent and Reason a) A car starts at -30 m and is moving in the negative direction at a speed of 10 m/s. The car is slowing down at +6.0 m/s 2. b) Initial Position: -30 m Initial Velocity: -10 m/s Acceleration: 6.0 m/s The car is slowing down.

2 PUM Physics II - Kinematics Lesson 16 Solutions Page 2 of 7 c) -x v 1 v 2 v 3 v 4 Δv 12 Δv 23 Δv 34 d) Velocity vs. Time e) Position vs. Time When the car stops I expect to see the slope of the line at that point to be exactly horizontal. v(t) = (-10 m/s) + (6 m/s 2 )t 0 m/s = (-10 m/s) + (6 m/s 2 )t t = 1.7 s

3 PUM Physics II - Kinematics Lesson 16 Solutions Page 3 of 7 x(t) = (-30 m) + (-10 m/s)t + (3 m/s 2 )t 2 x = (-30 m) + (-10 m/s)(1.7 s) + (3 m/s 2 )(1.7 s) 2 = m Traveled 8.3 m from initial position before stopping. f) 1.7 s see above. If the acceleration does not change the car will eventually reverse direction and go backwards. This isn t really realistic for the breaks in a car since the brakes bring the car to a completely stop Represent and Reason The graph is missing units. I ll assume the units are kilometers and minutes since the questions have those units. Motion Diagrams: Rocket A: -x Rocket B: -x a) Rocket A: (70 km 140 km)/(11 min) = -6.4 km/min 6.4 km/min Rocket B: (20 km 140 km)/(11 min) = km/min 10.9 km/min b) x(t) A = (140 km) + (-6.4 km/min)t x(t) B = (140 km) + (-10.9 km/min)t c) Rocket A after 6 min (looking at the graph): 140 km 100 km = 40 km Check with equation of motion: x A = (140 km) + (-6.4 km/min)(6 min) x A = km 140 km km = 38.4 km

4 PUM Physics II - Kinematics Lesson 16 Solutions Page 4 of 7 d) Rocket B to 60 km: Looks to be about 7 min Check with equation of motion: (60 km) = (140 km) + (-10.9 km/min)t t = (-80 km)/(-10.9 km/min) = 7.3 min e) Rocket A to 60 km: (60 km) = (140 km) + (-6.4 km/min)t t = (-80 km)/(-6.4 km/min) = 12.5 min Rocket B gets there 5.2 min sooner f) Zero 16.5 Represent and Reason a) Sketch and translate: Draw a sketch of the initial situation and choose a coordinate system to describe the motion of both bikes. Represent physically: Draw a motion diagram for each bike. 0 You 12 m/s 8 m/s 400 m West +x Your friend: +x Represent mathematically: Construct equations that describe the positions of each bicycle as a function of time x(t). x(t) You = (400 m) + (8.0 m/s)t x(t) Friend = (12 m/s)t

5 PUM Physics II - Kinematics Lesson 16 Solutions Page 5 of 7 Solve and evaluate: Use the equations to determine when the bicycles are at the same position. Does your result make intuitive sense? x(t) You = x(t) Friend (400 m) + (8.0 m/s)t = (12 m/s)t t = (400 m)/(4.0 m/s) = 100 s This makes sense. It will take about 100 s for my friend to catch up with me. b) Position vs. Time (Red is your friend, you are blue) Velocity vs. Time (Red is your friend, you are blue) c) Yes.

6 PUM Physics II - Kinematics Lesson 16 Solutions Page 6 of 7 d) Yes the two lines intersect at 100 s this is the time when the two bicycles were predicted to be at the same location. e) Done. f) The signs are all consistent with my coordinate system. The bikes are moving in the positive direction with positive velocities. I start out 400 m whereas my friend starts out at 0 m Real World Application a) From 16.2 v o = 24 m/s or about 54 mi/hr. It took 55 m or about 180 ft to stop according to our calculation. According to the chart the stopping distance should be between 359 to 268 ft. This discrepancy. might be because the reaction time at night is much slower than during the day. You could be tired and it is dark, making it harder to see. The reaction distance is much greater than what we calculated, between ft. We calculated 19 m or about 62 ft. b) Assuming the same reaction time, as the speed of the car increases, the distance that the car covers over that same time interval increases. c) 20 mi/hr = 29.3 ft/s Reaction Distance: 44 ft (44 ft)/(29.3 ft/s) = 1.5 s Check: 30 mi/hr = 44 ft/s (66 ft)/(44 ft/s) = 1.5 s NJ MVC probably selects this much greater time interval to warn people to drive slower at night. d) At 70 mi/hr (31.3 m/s) it takes 310 ft (94.5 m) to break. v 2 = v o 2 + 2a(x-x o ) a = (v 2 v o 2 )/[2(x-x o )] a = (-31.3 m/s) 2 /[2(94.5 m)] = -5.2 m/s 2 Acceleration in 16.2 was -8.0 m/s 2

7 PUM Physics II - Kinematics Lesson 16 Solutions Page 7 of Reason a) 8 min = 480 s (3*10 8 m/s)(480 s) = 1.4*10 11 m We can consider both objects to be point like in this case because the distance between them is so great. Earth Radius: 6.4*10 6 m Sun Radius: 7.0*10 8 m b) We will see the explosion when the light gets to us. 3.97*10 13 km = 3.97*10 16 m (3.97*10 16 m)/(3*10 8 m/s) = 1.323*10 8 s = 4.2 years