Chapter 1 Review Questions Knowledge 1. Uniorm motion occurs when an object travels in a straight line at a constant speed.. Four scalar quantities include time, distance, speed, and mass. 3. Four vector quantities include displacement, velocity, acceleration, and orce. 4. Vector quantities have both magnitude and direction. Scalar quantities have only magnitude. This is why the magnitude o a vector quantity is a scalar. 5. a. 5. 4 h 3600 s 19 440 s 4 1.9 10 s 1 b. 900 s - 50 h or.50 10 h 3600 s c. 1500 m 1. 500 m d. 4. 75 km 4750 m 4.75 10 m e. 35. 5 m s 3600 s 18 km/h. 85 km h 4 m/s 3600 s 6. a. Velocity is a vector quantity describing a change in position over a speciied time. b. Velocity is the slope o a position time graph. c. d D 7. a. Acceleration is a vector quantity describing a change in velocity over a speciied time. b. Acceleration is the slope o a velocity time graph. c. a i 8. The reaction distance is the distance a vehicle travels while the driver is reacting to some condition or hazard. Since the vehicle is travelling with uniorm motion, the reaction distance is equal to the speed o the vehicle multiplied by the reaction time o the driver. Braking distance is the distance the vehicle travels rom the instant the driver s oot touches the brake pedal to the time the vehicle stops. Because the vehicle is decelerating, the calculation o the braking distance is a two-step process: step 1: Determine the time it takes or the vehicle to stop. step : Calculate the distance using one o the accelerated motion equations. Stopping distance is the sum o the reaction distance and the braking distance. Science 0: Unit B 3 Chapter and Unit Review Suggested Answers
9. An applied orce is an external orce applied to an object. I you were pulling a toboggan across the snow, the orce you apply through the rope is the applied orce on the toboggan. The orce o riction is a contact orce between two suraces that acts to oppose the motion o one surace past the other. To return to the example o pulling a toboggan across the snow, the orce o riction is the orce between the bottom o the toboggan and the snow that pushes in the opposite direction o the toboggan s motion. The net orce is the vector sum o all the orces acting on the toboggan. When a toboggan is being pulled across the snow, the net orce is the orce that remains once the orce o riction has been subtracted rom the applied orce. 10. Newton s irst law states that, in the absence o a net orce, an object in motion will tend to maintain its velocity and an object at rest will tend to remain at rest. Newton s irst law is sometimes called the law o inertia. Newton s irst law explains why an unrestrained passenger in a vehicle will continue to move in the original direction even though the vehicle may have been stopped abruptly. 11. Newton s second law states that an object will tend to accelerate in the direction o an unbalanced orce such F that a net. Newton s second law explains why a orce is required to accelerate a vehicle. m Applying Concepts 1. Starting Ending Measured Displacement (m) Time to Reach Ending (min) Calculated Average Velocity (km/h) corner o Metcal St. and Gloucester St. corner o Bay St. and Queen St. Peace Tower 600 m [N] 9.0 4.0 km/h [N] National Arts Centre 105 m [E].0.80 km/h [E] Ottawa Locks Conerence Centre Supreme Court o Canada corner o Laurier Ave. and Elgin St. corner o Lyon St. and Sparks St. corner o Kent St. and Gloucester St. 530 m [S] 11.0.89 km/h [S] 940 m [W] 19.0.97 km/h [W] 590 m [S] 1.0.95 km/h [S] Science 0: Unit B 4 Chapter and Unit Review Suggested Answers
13. Starting Average Velocity (km/h) Time to Reach Ending (min) Calculated Displacement (m) Ending corner o Metcal St. and Wellington St. 3. km/h [W] 7.0 3.7 10 m [W] The Currency Museum Supreme Court o Canada 3.9 km/h [E] 10.5 6.8 10 m [E] Ottawa Locks corner o Kent St. and Slater St. National Arts Centre 5.3 km/h [N] 5.0 4.4 10 m [N] 4.4 km/h [W] 8.0 5.9 10 m [W] Supreme Court o Canada corner o Bank St. and Queen St. Peace Tower 4.3 km/h [S].5 1.8 10 m [S] Capital Ino. Centre 14. a. Starting at the corner o Bay and Wellington, the distance to the Capital Inormation Centre is 75 m. The position o the Capital Inormation Centre is 75 m, east. The position is the better description because it also includes the direction o how to get there. b. 75 m[ E] + 75 m + 0. 75 km 0.00 h + 0. 75 km + 3. 63 km/h 1. 0 min 00 h v? The velocity is 3.63 km/h [E]. c. W E - + 1 cm 100m 1 75 m[ E] resultant 395 m[ E] 330 m[ W] The resultant displacement is 395 m [E]. Science 0: Unit B 5 Chapter and Unit Review Suggested Answers
d. Determine the average speed. 75 m + 330 m 1055 m 1055 m 1. 055 km 1. 0 min + 5. 0 min 17.0 min 17.0 min 83 h? 1. 055 km 0.83 h 3. 7 km/h The average speed is 3.7 km/h. Determine the average velocity. 75 m[ E ]+ 330 m[ W] ( + 75 m) + (- 330 m) + 395 m + 395 m + 0. 395 km D t 83 h v? 0. 395 km + ( 0.83 h) + 1. 39 km/h The average velocity is 1.39 km/h [E]. The average velocity has a smaller magnitude because it is based on a smaller value or displacement. This is due to the act that one displacement is east (positive direction) while the other is west (negative direction). 15. a. The displacement o the Supreme Court o Canada is 490 m [W] o the Peace Tower. b. The vector notation is dropped here. 490 m[ W] - 490 m v - 0. 490 km - 0. 490 km 3. 0 km/h[ W] - 3.0 km/h - 3. 0 km/h 163 h D t? 163 h 9.8 min It will take 9.8 min to walk this distance. Science 0: Unit B 6 Chapter and Unit Review Suggested Answers
c. v 3. 3 km/h[ S] - 3. 3 km/h 8. 0 min 8. 0 min 13 h? d D v (- 3. 3 km/h)(0.13 h) - 0. 440 km - 0. 440 km - 440 m 440 m[ S] Using the map, the inal position is very close to the corner o Slater and Metcale. i 16. a. 9.0 s a + 4. ( + 16 ) i. - + 4. + 16. 9.0 s + 1. 3 m/s i b. 45.0 s a + 4. 5 cm/s i ( + 0. 5 cm/s ) - + 4. 5 cm/s + 0. 5 cm/s 45.0 s - 0. 089 cm/s i c. 18.0 s a + 3. 5 m/s i ( + 3. 5 m/s ) - + 3. 5 m/s + 3. 5 m/s 18.0 s 0 Ê v + v i ˆ Á È + 4. 16. Í Î + 90 m Ê v Á + ( + ) ( 9. 0 s) i + v ˆ + ( + ) È + 4. 5 cm/s 0. 5 cm/s Í ( 45. 0 s) Î + 1. 1 10 cm v + 3. 5 m/s 18.0 s + 63 m Note: Since this graph describes uniorm motion, an accelerated motion equation is not required to determine displacement. i d. 1.8 h a + 90 km/h i ( + 60 km/h ) - + 90 km/h + 60 km/h 1.8 h -17 km/h Ê v + v i ˆ Á È + 90 km/h 60 km/h Í Î 1. 4 10 km + ( + ) ( 1. 8 h) Science 0: Unit B 7 Chapter and Unit Review Suggested Answers
17. W E - + reaction distance stopping distance braking distance v 96. 0 km/h v 96. 0 km/h v 0 18. a. The motion o the car in the section labelled reaction distance is uniorm motion. By Newton s irst law, the car is able to maintain its velocity in this section because the orces on the car are balanced. This motion will not last long because as soon as the driver pushes on the brake pedal, the orces on the car will become unbalanced. b. 96. 0 km/h[ E] 1000 1. 0 + 96.0 km m h v h 1.0 km 3600 s ( + 6. 6 m/s)(.0 s) + 6. 6 m/s 53. 3 m D t.0 s 53 m D d? The reaction distance is 53 m. 19. a. The motion o the car in the section labelled braking distance is negatively accelerated, or decelerated motion. The car had an initial velocity o 96.0 km/h, east, and a inal velocity o zero; thereore, the car was decelerating. Newton s laws explain this urther. The negative acceleration means that the acceleration vector was pointing in the negative (or westward) direction due to a net orce, which was also acting in the negative (or westward) direction. This net orce was negative because the orces that caused the braking must push opposite to the original direction o the car, which was positive. b. v i + 6.6 m/s, v 0, a - 3.45 m/s,?,? First, determine the change in time. v - v i a v v a - i v - i a 0 - ( 6. 6 m/s) (- 3. 45 m/s ) 7. 79 468 599 s Next, determine the braking distance. ( v + v i ) Ê 6. 6 m/s + 0ˆ Á 103. 059 581 3 m 103 m ( 7. 79 468 599 s) The braking distance is 103 m. Science 0: Unit B 8 Chapter and Unit Review Suggested Answers
0. a. + stopping reacting braking ( 53. 3 m ) + ( 103. 059 581 3 m) 156 m The stopping distance is 156 m. b. I this emergency stopping manoeuvre had happened at night, the driver would not have been able to stop in time. As the previous answer indicates, the driver needed 156 m to stop the vehicle, so even with the added illumination o the HID headlights, the driver would not have been able to stop. c. The speed limit is not the speed that you are supposed to travel. The speed limit is the maximum speed or a vehicle under optimum conditions. The act that this happened at night (reducing visibility) and the act that the road was slippery (reducing the ability to decelerate) meant that the driver should have been travelling at a velocity less than 96 km/h. Science 0: Unit B 9 Chapter and Unit Review Suggested Answers