Kinematics of Motion

1. ---------------------------- “Time is defined so that motion looks simple” -Blaise Pascal ---------------- Unit 2 The Kinematics of Motion Introduction How do bodies move? Some bodies may move along a straight line at a maintained rate, for example the bride walking along the aisle towards the altar. Cars on the other hand follow the road which is a series of curves and straight lines and sometimes are driven slow or fast. Falling bodies usually follow a straight vertical line and simply do it fast and faster. But when a stone is thrown straight forward it does not follow a straight line, how could that be explained? The topic of this unit is kinematics of motion. You shall learn to describe how bodies move using the ideas of distance, displacement, speed, velocity, time and acceleration. Objectives: At the end of the unit, the student should be able to: 1.Describe the motion exhibited by a body in terms of its speed, distance traveled and time 2.Calculate the speed, velocity, distance and displacement of a moving body and time during which the motion may take place; 3.Calculate the acceleration of a moving body. 4.Distinguish a body moving with constant velocity from another with changing velocity; 5.Recognize the important features of free fall motion and give examples 6.Account for the motion of projectiles from two-dimensions 7.Determine the factors that affect the acceleration of a body in uniform circular motion. Glossary Particle - a model of a moving body whose position can be described by locating one point so that effects of rotation and change of shape could be disregarded. Frame of reference - a coordinate system plus a time scale. Speed - distance travelled divided by the time it took to travel the distance. Velocity - refers total displacement divided by the time it took for the displacement to occur. Free falling body - a body that is moving freely under the influence of gravity. Distance - refers to the length of path travelled Displacement - the measure of the change in position Workbook in Mechanics & Heat Page 11

2. 2.1.Frame of reference; Distance, Displacement, Average Speed and Average Velocity A body may be seen at rest with respect to one person but moving to another or depending on the frame of reference used. This could be the ground, a post, a corner of a room or the top of a building. When a man walks 100 m it may mean that the length of the path travelled from the first position (origin) to the final position is 100 m. In Figure 1, the path could be one straight line (path1)or a curve (path 2) or even two straight lines (path3)or more lines/curves such that the man is back to his original position. The length of this path is called distance. In which case is the man farthest from the original position? The illustration shows that by following path 1 that is one straight line the man shall be farthest from his original position. So how can we clearly differentiate the motion the man makes when he travels the same distance but follow various paths? Point of origin O Path 1 Path 2 Path 3 Figure 1. A man walks 100 m in different ways as shown in paths 1, 2 & 3. There are cases when the distance travelled is not enough information about the motion because it cannot tell exactly where the body is after the event. This could be addressed by taking into consideration the positions of the body before and after the event. Let us choose a post as point of reference for the man (posts usually are more permanent than a car by the road). Consider for this case that the man walks only along the straight road. Set the position of the post as 0 so that points located along the line (referred as x-axis and to the right of the post are positive; negative when located to the left of the post. Call the first position of the man at x1 as A and his final position at x2 as B. When the man moves from x1 tox2, he is said to have undergone a certain displacement. B – A = 1 2 x x x    x O A( B ( For example, a man standing by the road (frame 1) sees the driver and passengers moving relative to the road while the driver of a bus (frame2) sees his passenger seated at his back to be at rest. post 1x ) 2x ) Fig.2. Positions of a particle at two times during its travel Workbook in Mechanics & Heat Page 12

3. at 8:30:02 s, and sometime later at 8:30:06 s, it was seen at 12 m to the right from the same post (x2 = 12 m). The change of position or displacement is the vector from point A to point B along the road (or x). Thus the displacement ∆x= (x2–x1). ∆x is positive where x2 is to the right of x1(Fig. 2); ∆x is negative where x2 is to the left of x1.(Fig. 3). The vector ∆x is always drawn from the tip of vector x1 to the tip of vector x2. As shown in Figure 2, the man was first seen 4m to the right from the post(x1 = 4 m) Eq. 2.1 x2 B A Δx= x2 - x1 x1 Fig 3. The displacement (∆x) is negative. The time the man was spotted at position A is different from the concept of time it took the man to move from A to B. This concept is called time interval. Time interval from t1 to t2, is defined as walk in 4.0 s, it is obvious that he was faster when he runs because it took him less time to travel the same distance. But suppose it took him 4.0 s s to travel 12.0 m, in which case is he faster: 8.0 m for 4 s or 12.0 m for 4.0 s? Well of course you will say the second because you travel a longer distance in the same time. But then again if the man takes 2.5 s to travel 5.0 m and in another instance travel 100 m in 4.0 s, in which case is he faster? Is it obvious? Which concept will tell you when a man is faster, time or distance? If we define average speed as total distance divided by the time interval, that is, ???????????? =????????????? then we can determine the length of path travelled per unit of time. Walking 8.0 m in 4.0s gives a speed of 2.0 m/s; running 12.0 m in 4.0 s gives a speed of 3.0 m/s while 5.0 m in 2.5 s is also 2.0 m/s. Now you can compare motion of several bodies through an examination of their speeds, not distance nor time. A more complete information is given by the average velocity defined as the total displacement divided by the time interval or ∆t = t2 - t1. Eq. 2.2 Now suppose the man runs instead of walk? If he runs the 8.0 m in 2.5 s instead of ???????????? ,   x x x    Vaverage = Eq. 2.3 2 1 vx   t t t 2 1 Workbook in Mechanics & Heat Page 13

4. For the example given in Fig. 2 m m v x 2 6  In Fig.3, this value is -2 m/s, since the body is moving to the left. Directions can also be given in terms of geographical terminologies: East, West, South and North. In some cases, specifying the direction of motion is necessary; for example by asking bystanders the direction of the suspected gunman, policemen would have a better chance of catching up. If the direction is not specified, the speed and distance may be adequate. Example What is the average velocity of the rider on his bike a)if he travels 1.20 km East from the house in 2.0 minutes; b)if he travels 5.0 km West from the house in 8 minutes c)if he travels 1.20 km East from the house in 2.0 minutes, proceeds 5.0 km East in 8.0 minutes, then turns around and travels to a spot 0.5 km West from the house in 10 minutes. d)What is the speed of the rider in the 20 minutes of travelling (in case c)?  12 4 in the direction of ∆x (to the right). m    2 s s s Solution: a) If we let x1 = 0 at t1 = 0 s, then x2 = 1.20 km East at t2 = 2.0 mins = 120 s   s ) 0 120 (  b)∆? = 5.0 ?????? = 5,000 ?????; ???∆? = 8.0min = 480 ?, then 480 ? West s Sometimes this is given as -10.4 m/s. The negative sign suggests that the direction is to the left or simply towards West. c) If we let x0 = 0 km at t0 = 0; x1 = 1.20 km = 1,200 m at t1 = 2.0 min = 120 s; x2 = 6.20km = 6,200 m at time t2 = 10 min = 600 s; x3 = -0.5 km = - 500 m at t3 = 20 mins = 1,200 s. The total displacement for the entire trip is (x3- x0) = -500 m - 0 = - 500 m, in the total time interval of 20 min = 1,200 s. (Positions east of the house are positive and positions west of the house, negative) The average velocity for the entire trip is s s 200 , 1 The negative value indicates that the direction of the average velocity is towards West from the house.  1200 0 m m East m    10 vx East s ?? = 5,000 ????? m = 10.4    500 m m ;    . 0 42 ave v x Workbook in Mechanics & Heat Page 14

5. d)The total distance traveled by the rider is 12.9 km = 12,900 m in 20 min = 1,200 s so that the speed of the rider is 10.7 m/s. Take note that the average velocity and speed for the same event need not be the same. These values are equal only when the body is moving in one straight line without turning back. Sample Problem Two automobiles are 150 kilometers apart and traveling toward each other. One automobile is moving at 60km/h and the other is moving at 40km/h mph. In how many hours will they meet? Answer: 1.5 hours 2.2 Instantaneous Velocity To detect irresponsible drivers, traffic enforces use the cctv/camera to gather information about passing vehicles. The data give the speeds of the vehicles at the instances that pictures were taken. In this case the data given are velocities or speed at identified moments. These are referred to as instantaneous velocities or speeds. This answers the question “what is the velocity/speed right now?” The consequence of giving the instantaneous velocity or speed is to make known the details of the motion which are not implicit with average values. For example, a bus travelled 120 km in 3 hrs. We can only tell from this information the average speed of 40 km/hr but nothing is known at the beginning or one hour after the start or anytime in its travel, or whether it is running faster or slower than 40 km/hr or if there was a moment of rest and where. So if there is no mention of average value, the mention of speed or velocity in this book shall mean the instantaneous value. A model walking down the ramp tries as much as possible to keep her pace constant meaning her speed is constant and that suggests that she takes equal length of steps for equal intervals of time. If she maintains a forward direction, we can say that she exhibits uniform linear motion which implies constant Instantaneous velocity. As instantaneous velocity is maintained for some time, the average velocity matches with the instantaneous value. Remember velocity is magnitude and direction. As the model prepares to return she slows down and velocity changes. As she turns around for the return trip velocity also changes. Some changes their velocities so quickly while others just take their time and changes slowly. Workbook in Mechanics & Heat Page 15

6. 2.3 cannot help but move slow at the lobby and move fast where traffic is not heavy. Whenever a moving body undergoes changes in velocity, it is said to undergo acceleration. Let us define average acceleration as the change in velocity divided by the time interval that the change takes place. - A vector quantity. - The unit of acceleration is m/s2. - A constant acceleration of 2 m/ s2 means that the velocity of the body increases by 2 m/s every second. - A body which is moving to the right with an acceleration of 2 m/ s2 will have an increased velocity of 2 m/s every second. - A body which is moving to the left with an acceleration of 2 m/ s2 will reduce its velocity by 2 m/s every second until its velocity becomes zero. If it continues with that acceleration, it will start moving to the right with increasing speed. Questions: 1. A ball hits the floor vertically downward, bounces to a height of one meter, falls, and hits the floor again. Is the ∆y between the two impacts equal to zero, one, or two meters? Is the ∆x between the two impacts equal to zero, one, or two meters? (∆y refers to the vertical component of displacement; ∆x, to the horizontal component of displacement) 2. A contestant swam two lengths in a 50-m pool in 60.0 s. What is his average speed? What is his average velocity? Are the two values the same? Average Acceleration At the start and end of a class period, in moving from one room to another, students ? =Δ? Δ? = ?2−?1 Δ? Eq. 2.4 A A body moving to the right at 5 m/s with acceleration of 2 m/s2.. A body moving to the left (at - 5 m/s) with acceleration of -2 m/s2. B Positive Negative A body moving to the right at 5 m/s with acceleration of .- 2 m/s2. C Negative Positive A body moving to the left at - 5 m/s with acceleration of .2 m/s2. D Positive x,y plane Positive x Velocity Acceleration Description of the motion Positive/negative Positive/negative Velocity in that direction increases Velocity in that direction decreases Velocity in that direction decreases Only velocity along x is affected; vy remains the same A body thrown diagonally up with an acceleration of ax = 2 m/s2. Workbook in Mechanics & Heat Page 16

7. E Positive x,y plane Negative y Only velocity along y is affected; vx remains the same A body thrown diagonally up with an acceleration of ay= -2 m/s2. Table 1.Description of motion with acceleration. See examples below. Note: Positive here means to the right or upward while negative means to the left or downward. Example A ball rolls up a slope. At the end of three seconds its velocity is 20 m/s; at the end of eight seconds its velocity is 0. What is the average acceleration from the third to the eighth second? Solution: ? =Δ? 8 ?−3 ? 5 ? The average acceleration is -4 m/s2. 2.4 Uniformly Accelerated Motion and free-falling bodies. A special case of accelerated motion is the straight-line motion with constant acceleration. This is also called the uniformly accelerated motion. In this case, the velocity changes at the same rate throughout the motion. One example of this motion is a freely- falling body. It has a constant acceleration if the effects ofthe air are not important. Consider the following trends given in Table 2. Table 2. Data and results taken from a ball rolled down an inclined plane. (Distances were measured in points, a unit that equals 29/30 mm.) Time t(equal intervals) 0 0 1 1 2 4 3 9 4 16 5 25 6 36 7 49 8 64 Source: As described by Stillman Drake in, “The Role of Music in Galileo’s Experiments,” Scientific American, June 1975, p.98. Referring to Table 2 that shows the distances travelled by a ball on an inclined plane at several times as indicated.: Δ?=?2−?1 ?2−?1=0 ?/?− 20 ?/? = −20?/? = −4 ?/?2 t2 Distance x(points) 0 33 130 298 526 824 1192 1620 2104 x/t2 33.0 32.5 33.1 32.9 33.0 33.1 33.1 32.9 Workbook in Mechanics & Heat Page 17

8. 1.Is the distance travelled in the first second equal to the distance travelled in the 2nd second? That of the 3rd second? 2.What do you notice about the distances travelled by the ball in the succeeding equal time intervals? 3.How many times greater is the distance travelled in the 2nd second than in the 1st second? 4.How many times greater is the distance travelled in the 3rd second than in the 1st second? 5.What do you notice about the values of x/t2 found in the last column of table 1? Can we say that x increases with the same proportion as t or with the same proportion as t2? This trend is seen when the ball referred above is made to go down a plane inclined at an angle from a leveled table. As the angle of inclination is increased however, the x/t2 values are also bigger including angles almost 900. This result could be extended to bodies that are dropped (the angle of inclination is equivalent to 900). Where the effect of air resistance is negligible. It can be deduced from these results that objects fall in the same way when dropped from a height. x  2t The relation may be written as an equality by putting a proportionality constant. Let us find out what this k value is. From the definition of average velocity, the displacement is v v    x =kt2 where k does not depend on the nature of the falling object. Eq.2.5       Δx = (v ave) Δt =  2 1 t t 2 1 2      v v From the definition of average acceleration         2 1 a thus v v a t t 2 1 2 1  t t 2 1 In the case of v1 = 0 at t1 = 0 and x1 = 0, and v2 = a t2   v   t     2 Δx= x2 = 2 Since the subscript (2) is arbitrary, we can just drop it off. If a t2 is substituted in the former equation   2 2   2t   at a  x =   t 2 at x  with x = kt2, Comparing 2 a = 2k Eq. 2.6 Workbook in Mechanics & Heat Page 18

9. All free falling bodies fall with the same acceleration called the acceleration of gravity and denoted by g. Its standard value is 9.80665 m/s2 or approximately 9.81 m/s2 on the surface of the earth. It is slightly lower at higher altitudes and below the surface. For this book, the value 9.8 m/s2 shall be used. On the surface of the moon this acceleration is only about 1.6 m/s2 QUESTIONS: 1.A body that is thrown up leaves your hand with an initial velocity. Being under the same influence of gravity and if air resistance is negligible, the body’s acceleration is approximately 9.81 m/s2. While the body is going up, is it free falling? Justify your answer. 2. A ball thrown vertically upward rises to a maximum height and then falls to the ground. What are the ball’s velocity and acceleration at the instant it reaches its maximum height? Justify your answer. 2.5 Projectile Motion A body that is dropped, or thrown vertically down or up, falls to the ground as a free falling body. Suppose the body is instead thrown to the side or diagonally, will it fall to the ground? Where? Projectile motion refers to the motion of an object thrown into the air horizontally or at an angle from the horizontal. Examples: a volleyball tossed by a player to the opponents’ side; a man who jumps from the bus that is moving forward or relief goods dropped from a flying helicopter; and fireworks and water fountains. Projectiles follow a curved path. This could be explained by the following: A projectile is kept under control by two independent motions, which work independently together to create a precise mathematical curve which we now call a parabolic curve. The projectile’svertical motion is influenced by the force of gravity pulling the object towards the earth giving it a downward acceleration of 9.8 m/s2. The projectile’s horizontal forward motion is uniform implying that horizontal velocity is constant Workbook in Mechanics & Heat Page 19

10. In the absence of air resistance, all objects fall with the same uniform acceleration. Thus, two objects of different sizes and weights, dropped from the same height, will hit the ground at the same time. Suppose an object is thrown horizontally at the same time and from the same height from where another object is dropped. Which will hit ground first? Some projectiles are thrown fast, others are slow; but their acceleration due to gravity are still the same. No matter how large the horizontal velocity is, the downward pull of gravity is always the same. If an object, A, is thrown horizontally faster than another object, say B, from the same height, which of the two will hit ground first? The horizontal distance covered by a projectile in its flight is called the range. The range of an object that is dropped is zero, can you tell why? For the two objects, A and B, mentioned above, whose range will be smaller? All these questions can be answered from the analysis of projectile motion. Example Consider this: A ball is tossed into the air with a velocity of 10 m/s at an angle of 30 degrees with respect to the ground. That is its initial velocity which can be analyzed as having a horizontal component of 8.7 m/s and a vertical (upward) component of 5m/s. voy v0x = vocos 300 = 10 (.87) = 8.7 m/s vo voy = vo sin 300 = 10 (0.5) = 5.0 m/s Fig. 5.The components of the velocity of a projectile. In the absence of air resistance and any horizontal force, the ball will (a) maintain its horizontal velocity (b) At the same time it will go up but not forever. Just like an object that is thrown vertically upward it will slow down, stop momentarily upon reaching a particular height that depends on how fast it is thrown up, and thereafter fall to the ground. In executing both motions at the same time (one motion being independent of the other) the ball will have both a horizontal displacement and a vertical displacement. Its path would be something like that shown in the figure below. 300 v0x vo Δy to=0 Δx t Fig. 6.The path of a projectile. Workbook in Mechanics & Heat Page 20

11. (c )The range Δx=vox t (vox is the x component of initial velocity and t is the time interval from start to finish, as shown) (d) The vertical displacement Δy = voyt + gt2/2. (e) In the case of the ball mentioned earlier, the ball will take 0.5 s to go up and another 0.5 s to come down, a total of 1.0 s for the entire flight. Can you check this? (f) In the absence of air resistance, the ball can only rise as far as 1.25 m in 0.5 s and reach a horizontal distance of 8.7 m in 1.0 s. If the horizontal component vox were 17.4 m/s for the same v0y, the range would have been 17.4 m. Differences in the values of Δx and Δy largely depend on the values of vox and voy respectively. If the ball were tossed horizontally instead (like a stone thrown into the water from a bridge), it will no longer go up so its flight may look like the one shown below. In this case voy is zero and vy continuously increase according to the rate of g. The range depends on the value of vo. Can you now answers the questions previously asked? Example A farmer climbed a coconut tree and 5.0 m above the ground, he throws a coconut horizontally to a basket 6.0 m farther away from the foot of the tree. How fast must the farmer throw the coconut? Solution: Take note, the farmer throws the coconut horizontally so vo =vox = ∆? Time ?is also the time for the coconut to fall the vertical height of 5.0 m. In free fall and with an acceleration of 9.81 m/s2, it shall take the nut 1.01 s to fall the 5.0 m height. voy = 0; vy = gt; also ∆? = ????? ? = (Take note that the time to go up the height Δy is half of t). v0 Δy Δx Fig. 6.Path of a projectile thrown horizontally. ? = 6.0 ? ? ?? 2?; thus ?2= 2 (5.0 ?) ; ? = 1.01 ? 9.8? ?2 vo =vox = 6.0 ? 1.01 ? = 5.94 m/s Workbook in Mechanics & Heat Page 21

12. Example A man dropped himself off from a bus while it was running forward at 8 m/s. If the man touched ground in 0.3 s, how high is the bus from the ground? Solution: Take note that voy = 0; ?????= ∆? = ????? ? = Sample Problem At a stop light, a truck traveling at 15 m/s passes a car as it starts from rest. The truck travels at constant velocity and the car accelerates at 3 m/s2. How much time does the car take to catch up to the truck? Ans: 10 seconds 2.6. Circular Motion Uniform circular motion is motion in a circle with a constant speed. When moving in a circle, an object travels a distance equal to the perimeter (or circumference) of the circle, thus Average speed = circumference For objects moving around circles of different radii in the same period (T), the object traversing the circle of larger radius must be traveling with the greatest speed. The average speed and the radius of the circular path are directly proportional. Objects moving in uniform circular motion will have a constant speed. But does this mean that they will have a constant velocity? Fig. 7.The velocity vector of a body in circular motion. The direction of the velocity vector at every instant is in a direction tangent to the circle. Therefore, the direction of motion of an object moving in a circle is continuously changing. While the car’s speed is constant but its velocity is changing, acceleration is happening. ?? 2 and vy = gt ?? 2? = (gt) 2? = ??2 2 = (9.8? ?2 )(0.3 ?)2 2 = 0.44 ? =2?? ? time Workbook in Mechanics & Heat Page 22

13. The acceleration of a body moving with constant speed in the circle has magnitude ?2 it is proportional to the square of the speed and inversely with the radius of the circular path. Acceleration is a vector quantity and in this case its direction is towards the center of the circle. Because of its direction, it is also called the centripetal acceleration (ac). The relationship between ac, linear speed v and radius implies that on the same circular road, increasing the speed increases the acceleration that is if the speed is doubled, the acceleration increases four times and reducing the speed to ½ reduces the acceleration to ¼ the original value. Furthermore, while traveling the zigzag road with the same speed, the greater the radius the lesser the acceleration of the moving car. An increase in the radius by 2x reduces the acceleration to ½ the original value while the speed is maintained. Example A car has tires of 32.5 cm radius. If the car moves through a linear distance of 200 m in 10 s on the road with the tires turning, (a) how many turns have the tires made? (b) What is the linear speed of the car in km/hr? (c) Calculate the centripetal acceleration of the tire. Solution: (a)N = ????????????? (b)v = ? 10 ?? 1 ?? 3,600 ? (c)a = ?2 ? = ? ? = ∆? ? ; ?? = ?? = ? = ? ∆? = ?? vb ? ; ? ?? = ?2 a =∆? ? = ?? ? Fig. 8. The acceleration of a body in circular motion, ? ; that is, 200 ? 2?(.325 ?) = 98 ????? ????????????? = ? = 200 ? 0.20 ?? 10 ? = = 72 km/hr 2 ? = 20 ? ? 0.325 ? = 1,230 ? ?2 Example Calculate the centripetal acceleration of the earth as it makes 1 turn about the sun in 365 days and it is approximately 1.5 x 1011 m from the sun. Assume that the orbit is circular. Solution: ? = 365 ????? 24?? ?? ac =?2 ?2 2?(1.5 ? 1011?) = 3 x 104 m/s ??? 3,6??? ? = 3 ? 1042 1.5 ? 1011 = 0.006 ? Workbook in Mechanics & Heat Page 23

14. This value is very small compared with the acceleration due to gravity that is why we are not aware that we are experiencing it. Sample Problem 1. An object is moving on a circular path of radius π meters at a constant speed of 4.0m/s. Find the time required for one revolution. Answer: ?2 2. A car rounds a 20-m radius curve at 10m/s. Find the magnitude of its acceleration. Answer: 5.0 m/s2 Think about it: 1. Is it possible for an object (a) to be slowing down while its acceleration is increasing in magnitude; (b) to be speeding up while its acceleration is decreasing? In each case, explain your reasoning. 2. You throw a baseball straight up in the air so that it rises to a maximum height much greater than your height. Is the magnitude of the acceleration greater while it is being thrown or after it leaves your hand? Explain. 2? Workbook in Mechanics & Heat Page 24

15. Problem Set #02(The Kinematics of Motion): 1. Starting from a pillar, you run 200 m east (the +x direction ) at an average speed of 5.0 m/s, and then run 280 m west at an average speed of 4.0 m/s to a post. Calculate (a) your average speed from pillar to post and (b) your average velocity from pillar to post. 2. A brick is dropped from the roof of a building. The brick strikes the ground in 2.50 s. You may ignore air resistance, so the brick is in free fall. (a) How tall, in meters, is the building? (b) What is the magnitude of the brick’s velocity just before it reaches the ground? 3. A small rock is thrown vertically upward with a speed of 18.0 m/s from the edge of the roof of a 30.0-m-tall building. The rock doesn’t hit the building on its way back down and lands in the street below. Air resistance can be neglected. (a) What is the speed of the rock just before it hits the street? (b) How much time elapses from when the rock is thrown until it hits the street? 4. The fastest measured pitched baseball left the pitcher’s hand at a speed of 45.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it? Workbook in Mechanics & Heat Page 25

16. Please cut neatly along the broken line Unit 2 – Kinematics of Motion Name Encircle the letter of your answer. Class Schedule Score 1. A plane traveling north at 200m/s turns and then travels south at 200m/s. The change in its velocity is: A. zero B. 200m/s north C. 200m/s south 2. A ball rolls up a slope. At the end of three seconds, its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. What is the average acceleration from the third to the eighth second? A. 2.5 cm/s2 B. 4.0 cm/s2 C. 5.0 cm/s2 D. 6.0 cm/s2 3. At time t = 0, a car has a velocity of 16 m/s. It slows down with an acceleration given by −0.50t, in m/s2 for t in seconds. It stops at t = A. 64 s B. 32 s C. 16 s 4. A racing car traveling with constant acceleration increases its speed from 10m/s to 50m/s over a distance of 60 m. How long does this take? A. 2.0 s B. 4.0 s C. 5.0 s D. 8.0 s 5. A ball is in free fall. Its acceleration is: A. downward during both ascent and descent B. downward during ascent and upward during descent C. upward during ascent and downward during descent D. upward during both ascent and descent E. downward at all times except at the very top, when it is zero 6. A car starts from rest and goes down a slope with a constant acceleration of 5 m/s2. After 5 s the car reaches the bottom of the hill. Its speed at the bottom of the hill, in meters per second, is: A. 1 B. 12.5 C. 25 D. 50 7. The acceleration of an object, starting from rest, is shown in the graph. Other than at t = 0, when is the velocity of the object equal to zero? A. During the interval from 1.0 s to 3.0 s B. At t = 3.5 s C. At t = 4.0 s D. At t = 5.0 s E. At no other time less than or equal to 5 s 8. A large cannon is fired from ground level over level ground at an angle of 30◦ above the horizontal. The muzzle speed is 980m/s. Neglecting air resistance, the projectile will travel what horizontal distance before striking the ground? A. 4.3km B. 8.5km C. 43km D. 85km 9. A projectile is fired from ground level over level ground with an initial velocity that has a vertical component of 20m/s and a horizontal component of 30m/s. Using g = 10m/s2, the distance from launching to landing points is: A. 40m B. 60m C. 80m 10. A bullet shot horizontally from a gun: A. strikes the ground much later than one dropped vertically from the same point at the same instant B. never strikes the ground C. strikes the ground at approximately the same time as one dropped vertically from the same point at the same instant D. travels in a straight line E. strikes the ground much sooner than one dropped from the same point at the same instant D. 400m/s north E. 400m/s south E. 6.67 cm/s2 D. 8.0 s E. 4.0 s E. The time cannot be calculated E. 160 E. 170km D. 120m E. 180m Workbook in Mechanics & Heat Page 26