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Name:_____________________ Regents Physics Date:________ Mr. Morgante UNIT 2B Kinematics – One Dimentional Motion Motion in One Dimension Motion is Relative Earth is moving at 100,000 km/hr orbital speed. Looks or feels like barely moving from our perspective. Frames of Reference: We must specify which frame of reference we’re using when describing the motion of a body. 1. In most cases, it will be the Earth itself. 2. When the body moves in a straight line. a. Choose any point on the line as a reference point. In kinematics and dynamics we will discuss two kinds of quantities (measurements): scalar: scalars have magnitude (a number value), but no direction e.g. time, mass, distance. Mass is a great example, since it has a number value (like 58 kg), but we don’t give it a direction (like “down”). vector: have magnitude and direction e.g. velocity, force, displacement. Force has a number value (like 37 N) and a direction (like “pushed to the left”). In kinematics we need to be able to have a way to describe the motion of the objects we will be studying, whether it’s a car or an atom. The most basic information you need to describe the motion of an object is… 1. position 2. time Speed and Velocity Speed Velocity Quality Scalar Vector speed and direction Units m/s, km/hr, cm/s 300 km/hr North, 5 m/s East Symbols V, V Average V Acceleration Def: The rate at which an objects speed is changing. The Symbol is a. 2 Units: m/s2 = meters/ seconds x seconds Acceleration Deceleration Increasing speed Sign (+) Decreasing Speed Sign (-) +a -a *Note: acceleration can be both (+) and (-) Straight Line Motion origin . -V +V Average Speed and Instantaneous Speed Average Speed det. over interval of time. Instantaneous speed –Speed determined from moment to moment ex: Wash. H.S. student Ex. Speedometer reading on I-87 at any time pulling out of parking lot “sick” Uniform and Accelerated Motion Demo Uniform Motion Accelerated Motion Speed and direction remained the same; CONSTANT VELOCITY At rest V0 = 0 Vi = 0 increases 3 Speed Important Note: Accelerated Motion *Definition: Motion with changing velocity * 3 cases that relate to vectors which include both magnitude & direction 1. Object changes direction in circular fashion (important when we do centripetal motion) or straight line with a sudden change in direction as shown below. 2. Object changes speed (this is magnitude) - accelerating speeding up to go through a yellow light 3. Object changes direction and speed Units of “sick” Acceleration Average speed and Instant Speed (continuous) *distance = average speed x time Units: d = v ∙ t Meters meters ∙ sec = meters sec Constant speed in a straight line d=v∙t Uniformly Accelerated Motion Simplest case: body moves in a straight line w/ constant acceleration. Speeds up (+) direction, slows down (-) direction. Galileo: showed that a body speeds up as it falls with an acceleration of 9.8 m/ s2 Gravitational pull produces acceleration Denoted by the letter g (little g) If a body is thrown upward, gravity (g) reduces speed. 4 Relating Acceleration, Speed and Time Body moving in straight line, constant acceleration. eq. a = V – V0 OR Vf -Vi t t Solving for V fields V=Vi + a∙t Vi= initial speed (sometimes you will see Vo) Vf= final speed uniform acceleration Average Speed& Distance Traveled During Constant Acceleration _ V= Vi +V _ 2 V= Vi∙t(Vi+a∙t) = 2Vi+at = Vi+ ½ a∙t 2 2 _ V= Vi + ½ at _ _ d=V∙t, OR t∙V= Vi∙t + ½ a∙t2 d= Vi∙t + ½ a∙t2 These are several formulas that are very useful when the acceleration is uniform. Do not use these equations if the acceleration is changing! These formulas are based on combinations of the basic velocity and acceleration formulas combined. Formula 1 A common problem is to figure out the velocity of an object after it has been accelerating for a certain time. You can take the basic acceleration formula and just solve it for final velocity to get… vf = vi + a∙t Example: I’m driving my Mustang at 61km/h when I notice that there is a school zone ahead. If I slam on the brakes for 6.7s and experience acceleration of 1.5m/s2, will I be under the 30km/h posted speed limit? First, change the initial velocity into m/s… vi = 61km/h = 16.9444444m/s Remember to keep all these extra digits on your calculator, but keep track of the actual sig digs. Keep in mind that since I was slowing down, the acceleration is negative. Now use the formula. This one is so common, you don’t even need to show how you manipulated it if you remembered it. 5 vf = vi + a∙t = 17m/s + (-1.5m/s2)(6.7s) = 6.894m/s = 24.81 km/h = 25 km/h Yep, I slowed down enough. Notice that I kept track of sig figs and changed it over to km/h at the end. Formula 2 We will also do problems where we need to calculate the displacement of an accelerating object after a time interval has passed and we know its initial velocity. d = vi ∙t + ½ a∙t2 Be careful with this formula. Only the time is squared in the last term, not acceleration and time. As a bonus, a lot of the time vi will be zero, which cancels out the first term and leaves you with… d = ½ a∙t2 Example: Occasionally the US Air Force calls me in to test fly their “birds”. A few weeks back I was flying along in my F-22 at 97m/s when I decide to kick in the afterburners for 12.3s. If the afterburners can generate enough thrust to accelerate the F-22 at 26m/s2, how far did I travel during that time? d = vi t + ½ at2 = (97m/s)(12.3s) +½ (26m/s2)(12.3s)2 = 3.2 x 103 m So I just traveled a little over 3 kilometers in about 12 seconds… not bad! Example: I am in a F-22 that is on the runway. From rest, I accelerate the plane at 3.9m/s2 for 9.5s to reach take off velocity. How long does the runway have to be? This is an example of a question where the initial velocity is zero (since I’m starting from rest), so… d = vi t + ½ at2 = ½ at2 = ½ (3.9m/s2)(9.5s)2 = 1.8 x 102 m 6 Formula 3 There is another formula related to the above that takes into account when we know the final velocity instead of the initial. d = vf ∙t - ½ a∙t2 Notice that the difference is the minus sign instead of addition, and final instead of initial velocity. We don’t use this in this class very often if at all. Formula 4 Another very useful formula is the following… vf2 = vi2 + 2a∙d Very handy when you have a question with both velocities, acceleration, and displacement. Don’t forget to do the square root at the very end if you are solving for a velocity, as the following example shows… Example: What is the final velocity of a car that starts at 42 m/s and accelerates at 3.78m/s2 for a distance of 12 m? vf2 = vi2 + 2ad = (42m/s)2 + 2 (3.78m/s2) (12m) 2 vf = 1855 Many people leave the answer like this, forgetting that this is velocity squared! vf = 43 m/s Formula 5 The last formula I’ll show you is probably the least used one. Only use it when you know for certain that the object has been going through a constant acceleration. Some people ignore that rule for this equation (although it applies to all these formulas). d = ½ (vf + vi) t Example: How far did a vehicle move if it started at 12 m/s and accelerated up to 47 m/s in a time of 34s? d = ½ (vf + vi) t = ½ (47 m/s + 12 m/s) (34s) Notice that it really doesn’t matter which order the velocities are added in. d = 1.0 x 103 m Summary So, how do you figure out which formula to use for a particular problem? Well, each of these formulas is missing one of the variables. Choose a formula that has the four variables in your question, and not the fifth missing variable. 7 he following table may help. Formula vf = vi + at d = vi t + ½ at2 d = vf t - ½ at2 vf2 = vi2 + 2ad d = ½ (vf + vi) t a vf X vi d X t X X X For example, if I had a question where I am given acceleration, displacement, and time, and asked to find initial velocity, I would choose the second formula. o It’s the only formula that doesn’t have final velocity, which I haven’t been given or asked for. Remember that for all of these formulas, you may be required to manipulate the formula to find the answer you are looking for. Always follow the rule of finding the formula that has all the knowns and unknown that you have, then manipulate it for your unknown, and solve. Example: What is the displacement of a car that starts at 10 m/s and accelerates at 1.89m/s2 and reaches a final speed of 32m/s? vf2 = vi2 + 2ad can you see why I choose this formula? 2 2 v vi (32m / s) 2 (10m / s) 2 d f 2a 2(1.89m / s 2 ) d = 2.4 x 102 m Freefall a = g = 9.81m/s2 ag = g = acceleration due to gravity Since gravity is just acceleration like any other, it can be used in any of the formulas that we have used so far. It is necessary to be particularly careful about using the correct sign (positive or negative) depending on the problem. Example: A ball is thrown up into the air at an initial velocity of 56.3m/s. What is its velocity after 4.52s have passed? 8 In the question the velocity upwards is positive, and I’ll keep it that way. That just means that I have to make sure that I use gravity as a negative number, since gravity always acts down. vf = vi + at = 56.3m/s + (-9.81m/s2)(4.52s) vf = 12.0 m/s This value is still positive, but smaller. The ball is slowing down as it rises into the air. Example: I throw a ball down off the top of a cliff so that it leaves my hand at 12m/s. How fast is it going 3.47 seconds later? In this question I stated a downward velocity as positive. I might as well stick with this, but that means I have defined down as positive. That means gravity will be positive as well. vf = vi + at = 12m/s + (9.81m/s2)(3.47s) vf = 46 m/s Here the number is getting bigger. It’s positive, but in this question I’ve defined down as positive, so it’s speeding up in the positive direction. Example: I throw up a ball at 56.3 m/s again. How fast is it going after 8.0s? We’re defining up as positive again. vf = vi + at = 56.3m/s + (-9.81m/s2)(8.0s) vf = -22 m/s Why did I get a negative answer? The ball reached its maximum height, where it stopped, and then started to fall down. Falling down means a negative velocity. maximum height There are a few rules like this that you have to keep track of. Let’s look at the way an object thrown up into the air moves. As the ball is going up… o It starts at the bottom at the maximum speed. o As it rises, it slows down. o It finally reaches its maximum height, where for a moment its velocity is zero. This is exactly half ways through the flight time. 9 As the ball is coming down… o The ball begins to speed up, but downwards. o When it reaches the same height that it started from, it will be going at the same speed as it was originally moving at. o It takes just as long to go up as it takes to come down. Example: I throw my ball up into the air (again) at a velocity of 56.3 m/s. a) How much time does it take to reach its maximum height? It reaches its maximum height when its velocity is zero. We’ll use that as the final velocity. Also, if we define up as positive, we need to remember to define down (like gravity) as negative. v vi a f t v v i 0 56.3m / s t f = 5.74s a 9.81m / s 2 b) How high does it go? It’s best to try to avoid using the number you calculated in part (a), since if you made a mistake, this answer will be wrong also. If you can’t avoid it, then go ahead and use it. vf2 = vi2 + 2ad 2 2 2 vf vi 0 56.3m / s = 1.62 x 102 m d 2 2a 2 9.81m / s c) How fast is it going when it reaches my hand again? Ignoring air resistance, it will be going as fast coming down as it was going up. The magnitude should be the same, but the direction will be different. You might have heard people in movies say how many gee’s they were experiencing. All this means is that they are comparing the acceleration they are feeling to regular gravity. So, right now, you are experiencing 1g… regular gravity. During lift-off the astronauts in the space shuttle experience about 4g’s. o That works out to about 39m/s2. Gravity on the moon is about 1.7m/s2 = 0.17g Speed & Velocity Notation for speed is V (This is all speed) Average Speed=V Units (MKS)= m/s Equation: V= d/t= distance / time = m/s Get out reference table & look at it. 10 If I have constant speed during an entire time period; it is said the object is in UNIFORM MOTION. Velocity - Time rate of change of an objects displacement. - Note: speed is a scalar and velocity is VECTOR. Speed & velocity are not interchangeable. Example: In the 1988 Summer Olympics, Florence Griffith-Joyner won the 100m race in 10.54s. Assuming the race is measured to the nearest 0.1m, what was her average velocity? Since the race is measured to the nearest 0.1m, our actual d = 100.0m d 100.0m v 9.488m / s t 10.54s Most of the time we talk about velocity in kilometers per hour. Graphs of Linear Motion d-t graph (distance –time) Independent variable (t)-time (sec) on x-axis Dependent variable (d) on y-axis d (m) what is the slope of this graph mean? Slope= ∆y/∆x = ∆d/∆t= V= speed what can you tell me about this objects motion? Answer: It’s constant! (or Uniform) t(s) d (m) 1 t(s) 3 Objects motion d (m) NOT MOVING Slope= ∆y/∆x= 0M/(3-1) =0 NO SLOPE B A t(s) Object’s motion= Not Constant Not Uniform Greater slope @ B Slope is NOT constant It is changing it’s increasing This object is accelerating 11 NOTE: Whenever you see a curved d-t graph it is either acc. or decelerating. depending upon the positive or negative slope of the graph. d (m) DECELERATING t(s) Billion $ Quest- How do I find instantaneous velocity? Ans: I take the slope of the point (time) that is tangent to the curve. Ex. d (m) y2 y1 Guaranteed question on quiz or test t(s) x1 x2 Acceleration Definition: Time of change of velocity A= ∆V where A= acceleration (m/s2) t t= time(s) units (m/s2) _ a = Average acceleration a= uniform or constant acceleration a (m/s2) t (s) How do I find the average velocity of an object’s acceleration uniformly? Vi = initial velocity Vf = final velocity If I have acceleration, I have a change in velocity! Vi < Vf Vi = 2 m/s Vf = 10 m/s V = average velocity when an object accelerates V = (Vi + Vf) / 2 = NOT ON REF. TABLE! KNOW IT – common sense NOTE: This is only valid when you have constant or uniform acceleration. 12 a (m/s2) a (m/s2) t (s) t (s) can’t use V = (Vi + Vf) / 2 for this Area Under Graph V (m/s) Area under the curve = L x W or Y * X = V x t (m/s) * (s) = (m) = distance traveled t(s) Area under v – t graph = distance traveled V (m/s) A = ½ b * h = dist. traveled t(s) We will utilize the various worksheets handed out in class for example problems. 13 14 15 Name____________________________________ Regents Physics Date_______ Mr. Morgante 1-D Motion Equation Note/Worksheet & Problem Solving Mechanics Equations Equation Variables Units Can be used to find… v = d t a = Δv t vf = v i + at d = v i t + ½ at2 v f2 = v i 2 + 2ad Problem solving methodology: 1. Write down all of the “knowns” in the equation. Examples are initial velocity (vi) and final velocity (vf) of an object. Note that some of the “knowns” may not actually be given, you have to know them. Examples are gravity (g) = 9.81 m/s2 and if you start “from rest” your initial velocity (vi) is zero! 2. Analyze your “knowns”, and understand what it is the problem wants you to determine. If the problem asks for final velocity, how many equations are there above that can be used? Be careful and look closely. 3. If you don’t understand what a subscript or superscript of an equation means, look on your cheat sheet, I mean your Reference Table. All the info is listed there right next to the equations. OVER 16 4. Plug and chug. It is recommended to do algebra first (w/o plugging in any numbers!) to arrange the equation so you can solve for your unknown. Using units analysis (a.k.a. canceling units) should help you to avoid mistakes. If you are solving for velocity and your units analysis yields m/s2, you obviously did something wrong so you should check your work. 5. All equations used should be listed and units should be used in every step of the equation. You should box your final answer, and your final answer should have units. Please follow as neat and orderly a procedure as possible……..if it is neat and orderly I can grade it easily and unnecessary mistakes will be avoided in your work and my grading. USE PENCIL SO YO CAN ERASE AND BE NEAT!!!! 6. This methodology can be used for all subsequent units we will cover. 17 Name:____________________ Regents Physics 1D-Motion Problem Scenarios Date:___________ Mr. Morgante Freefall (Vertical Motion): Ball thrown upward with an initial velocity: Vtop = 0 m/s g (negative value in equation) vi Things to consider: Sign convention is shown below. This is very, very important b/c it tells you what drirection something is moving in. +y (pos. initial velocity) -y (gravity always in this dir., & velocity on the way down) We almost always ignore air friction and we assume that the toss is almost perfectly straight up and down. The time it takes the ball to go up to its maximum height equals the time it takes for the ball to come down from the maximum height. This means there is SYMMETRY. If you analyze the various components of the ball on its way up (velocity, time, etc.), it will be equal to the components going down at the respective time. For instance, if vi=25 m/s, vf=25 m/s if the ball is caught at the same height it is thrown from. Gravity (g=9.81 m/s2) is working against the initial velocity upward thereby slowing the ball down until it reaches the top of its flight where the vertical velocity is zero. Gravity is therefore a negative value in the equations. At the top of the flight, it is as though the ball is being released from rest, so the vi =0m/s if the problem is analyzed from this point forward in time. Gravity is 18 working to speed the ball up when the ball is on its way down. Even though it is still a negative value in the equation, it does not make the ball slow down as is the case when the ball is going upward. The negative sign in the equation serves to tell the problem solver the direction the ball is moving in. Freefall Cont. Ball dropped from a cliff: g=9.81 m/s2 vi=0 m/s d Things to consider: Sign convention as discussed previously still applies. If you DROP something, it comes out of your hand with an initial vertical velocity that equals zero (vi=0 m/s). Gravity immediately starts to uniformly accelerate the object at 9.81 m/s2 so that the velocity increases in a straight with a constant slope if you were to graph velocity versus time. The ball’s velocity increases by 9.81 m/s every second! You get that from a=Δv. t 2 Where t= 1 second intervals and a=g=9.81 m/s . Since vi=0 m/s, Δv becomes vf. Fill out the table below to prove this is the case: Acceleration (a) in m/s2 Time (s) Velocity m/s You can also solve for unknowns such as distance or displacement. You can solve for d if only time is given since vi=0 m/s. These scenarios require practice that you will get in the class worksheets. 19 Horizontal 1-D Motion: Decelerating object: vi a Things to consider: Sign convention as follows: -v, -d or –a +v, +d or +a Here, the object has an initial velocity and the acceleration is opposing the motion of the object. This is called deceleration and would be inserted as a negative value in the equation chosen for the particular problem. Accelerating object: vi a Things to consider: Sign convention as follows is the same as above: -v, -d or –a +v, +d or +a Here, the object has an initial velocity and the acceleration is in the same direction as the motion of the object. This is called acceleration and would be inserted as a positive value in the equation chosen for the particular problem. 20 Inclined Plane 1-D Motion +y +x Ball released from rest on incline: Look at the coordinate system above, it is in alignment with the slope of the incline now! If the ball is on a frictionless plane, it will accelerate down the incline. It will accelerate because of gravity acting on the ball, thereby increasing its velocity. Ball starts from rest so vi=0 m/s. Ball can also start with a vi if I wanted to test your true understanding of the material. As the angle of the incline increases, the ball will accelerate faster b/c the acceleration component of gravity becomes greater! If we approached a 90° angle, the object would be in freefall and g=9.81 m/s2. If the angle is anything less than 90°, the acceleration will be less than g=9.81 m/s2, it will be a component of g=9.81 m/s2!!!!! The value of acceleration is usually given. Assume that acceleration is uniform, so v= vi + vf , v = d/t, and a=Δv 2 t Ball pushed/thrown upward on an incline: +y +x vi Have to have a vi here in order to have ball move up incline. vi is a positive value here. Gravity is working against ball to slow it down, hence g is a negative value here and is some component that is less than 9.81 m/s2. Assume that acceleration is uniform, so v= vi + vf , v = d/t, and a=Δv 2 t 21 NAME________________________________ DATE________ Regents Physics Mr. Morgante Kinematics Equations & Algebra Review Objective: Review equations, algebra and solution methods for kinematics 1. Write the equation for average speed:____________ Solve for: d t 2. Write the alternate equation for average speed: ___________ What special condition is required to use this equation?:_________________ 3. Write the equation for acceleration: ___________ Expand the acceleration formula using vf , vi ___________ Solve the expanded acceleration formula for vf : ____________ Solve the expanded acceleration formula for vi: ____________ Solve the expanded acceleration formula for t : ____________ 4. Write the equation for distance using vi , a & t: ____________ If vi = 0, how does the equation simplify? ____________ Solve the equation for a: ____________ 5. Write the equation for final speed (or velocity) using vi , a & d: ____________ Solve the equation for vi : ____________ Solve the equation for d: ____________ Solve the equation for a: ____________ 22 Name:______________________ Date:_______ Regents Physics Mr. Morgante Kinematics Problems 1. A ball, initially at rest at t = 0 seconds, rolls down a long incline. Since the object is on the incline, its acceleration will be less than 9.8 m/s2, although the acceleration will still be constant. If the ball has rolled 1 meter at t = 2 seconds, how far will it have rolled at t =4 seconds? A. 2 m B. 3 m C. 4 m D. 5 m E. 8 m 2. If an object is placed on a frictionless incline, it will slide down with a constant acceleration. However, because the object is not falling straight down, this acceleration will be less than the acceleration due to gravity. Assume specifically that an object is released from rest on a certain frictionless incline at time t = 0 and that the object slides a distance of 1 meter during the first second. If the object then keeps on sliding down this incline with the same acceleration, through what distance will it travel from t = 1 second to t = 2 seconds. A. 1 m B. 2 m C. 3 m D. 4 m E. 5 m 3. The picture below shows the pattern of dots that was made by a tape timer on a strip of ticker tape connected to a moving object. The tape is oriented so that the first dot that was made is at the left. Notice that vertical grid lines have been drawn to help you judge the distances. Which of the following best describes the motion of the object? A. B. C. D. E. The object has a constant velocity The object was speeding up with a constant acceleration The object is speeding up with a non-constant acceleration The object is slowing down with a constant acceleration The object is slowing down with a non-constant acceleration OVER 23 4. The picture below shows the pattern of dots made on ticker tapes connected to two different moving objects, I and II. Each timer was set to make the same number of dots per second and each tape is oriented so that the first dot that was made is at the left. Which of the statements below best compares the motion of the two objects? I II A. Object I has a smaller initial speed than object II and object I has a smaller average acceleration than object II B. Object I has a smaller initial speed than object II, but object I has a larger average acceleration than object II C. Object I has a larger initial speed than object II, but object I has a smaller average acceleration than object II D. Object I has a larger initial speed than object II and object I has a larger average acceleration than object II E. Object I has a smaller initial speed than object II but both objects have the same average accelerations. 5. Each of the three graphs above represents acceleration versus time for an object that already has a non-zero positive velocity at time t1. Which graphs show an object whose speed is increasing for the entire time interval between t1 and t2? I a II t1 A. B. C. D. E. t2 t a III t1 t2 t a t1 t2 t graph I, only graph II, only graphs I and II, only graphs I and III, only graphs I, II, and III OVER 24 6. Each of the graphs below shows the displacement, d, as a function of time for a different moving object. Which graph(s) show an object that is moving with a positive acceleration? d d t t t II I A. B. C. D. E. d d III t IV graph I, only graphs I and II, only graph I and III, only graph I and IV, only graph III and IV, only 0 t t 0 0 Distance Speed t0 t 0 <B> t t Speed t0 0 <C> 0 Speed Distance Speed t0 <A> Distance Distance 0 Speed Distance 7. Which pair of graphs below shows the distance traveled versus time and the speed versus time for an object uniformly accelerated from rest starting at time t = 0? t <D> t <E> OVER 25 Displacement Questions 8 to 11 : Each of these questions refers to the displacement versus time graph below. The graph describes the motion of a car that was initially heading north along a straight north-south road. Each of the five labeled intervals on the graph – A,B,C,D and E – lasts for the same amount of time. North is the positive direction. D C B E A Time 8. Over which interval(s) is the velocity of the object greater than zero? A. interval A, only B. interval B, only C. interval D, only D. intervals A, B, and C E. intervals A, B, C, and D 9. Over which intervals is there no change in the velocity of the object? A. intervals A and C, only B. intervals B, D, and E, only C. interval D, only D. interval B, only E. interval E, only 10. Over which interval does the object have the greatest average speed? A. interval A B. interval B C. interval C D. interval D E. interval E OVER 26 11. At the end of interval E, the car is A. back at its starting point B. further north of its starting point than it was at any other time during its trip C. north of its starting point, but not as far north as it was at other times during its trip. D. further south of its starting point than it was at any other time during its trip. E. south of its starting point, but not as far south as it was at other times during its trip. 12. The speed of an object one second after it has been dropped from rest out of a high office window on the planet Mitochondria is 8 m/s. What will its speed be 3 seconds after it was first dropped? A. 8 m/s B. 24 m/s C. 12 m/s D. 72 m/s E. 36 m/s 13. An object is thrown upward from the ground at sea level on Earth with an initial speed of 30 m/s. Which of the following is closest to the total time that the object will be in the air before it strikes the ground again? A. 2.5 seconds B. 6.0 seconds C. 3.0 seconds D. 5.0 seconds E. 30 seconds 14. A ball is thrown vertically up and is caught when it returns to the same vertical position from which it was thrown. The ball takes 3 seconds to reach its maximum height. For what total time interval is the ball in the air? Neglect air friction. A. between 3 seconds and 6 seconds B. 6 seconds C. longer than 6 seconds D. 9.8 seconds E. 19.6 seconds 27 Name:________________ Date:_________ Regents Physics Mr. Morgante Graphical Analysis I (Displacement vs. Time) – Remember Sign Convention Displacement (m), East Displacement (m), East vs. Time (s) 12 10 8 6 Series1 4 2 0 0 5 10 15 Time (s) 1. What is the displacement of the object from the datum or origin point (0,0) at: t=0.0 sec?_______________ t=3.0 sec?_______________ t=7.0 sec?_______________ t=10.0 sec?______________ t=13.0 sec?______________ 2. What is the speed of the object at: t=1.0 sec?______________ t=6.0 sec?______________ t=9.0 sec?______________ t=12.0 sec?_____________ 3. What is the object’s velocity from t=0 to: t=0.0 sec?_____________ t=3.0 sec?_____________ t=7.0 sec?_____________ 28 Name:________________ Date:_________ Regents Physics Mr. Morgante Graphical Analysis II (Velocity vs. Time) – Remember Sign Convention! Velocity (m/s), North vs. Time (s) Velocity (m/s), North 14 12 10 8 Series1 6 4 2 0 0 5 10 Time (s) 1. What was the speed of the object at t=7.0 s? 2. How far does the object travel in the first 4.0 seconds? 3. What was the value of the object’s acceleration at t=0.5 seconds? 4. The graph shows that the object has an acceleration 0.0 m/s2 at t=________ (Show your work below) 5. The object’s acceleration was directed opposite to its velocity at a time of t=_________ (Show your work below) Note: Remember the area under the curve for a v-t graph will yield the displacement!!!!!! Note#2: If you can do v-t and d-t graphs, you can figure a-t graphs out on a test or quiz I give!!!! 29 Name:_____________________ Regents Physics 1-D Motion Graphs Worksheet Date:_________ Mr. Morgante Sketch the general form of d-t, v-t and a-t graphs for the following 1-dimensional motion scenarios: 1. Tossing an object straight up in the air and then catching it at the same height it was thrown from. t d t v t a 2. An object moving at constant velocity in a positive direction (it has an initial velocity at t=0 sec). t d t v t a 3. An object that starts from rest and moves in a straight line in a positive horizontal direction with uniform acceleration. t d t v t a 4. Dropping an object from an arbitrary height where origin is the height dropped from. t d t v t a OVER 31 +y +x 5. An object that rolls down an inclined plane. t d t v t a +y +x 6. An object that rolls up an inclined plane and stops at the top of the incline. t d t v t a 7. An object that has an initial velocity in a positive horizontal direction undergoes deceleration to the point where it stops. t d t v t a 8. An object that has an initial velocity of 20 m/s in a positive horizontal direction and undergoes deceleration to the point where it has a final velocity of 10 m/s in a positive direction. t d t v t a 32 Name________________________________ Regents Physics Acceleration in x or y plane Show all work. Place answer in space provided. Date________ Mr. Morgante 1. A car is accelerated uniformly from rest to 40 m/s in 20 seconds. a) What is the magnitude and direction of the car’s acceleration? Sketch Calculations a)__________ b) What is the magnitude and direction of the car’s displacement? b)__________ c) What is the car’s velocity at 16 seconds? c)__________ 2. A baseball is thrown straight up (+y direction) at a velocity of v0y = +35 m/s. use magnitude of g = 10 m/s2 a)What is the ball’s velocity at t = 1.0 second? Sketch Calculations a)__________ b) What is the displacement of the ball at time t = 2.0 seconds? b)__________ 33 Name:_____________________ Regents Physics Date:_______________ Mr. Morgante Kinematics Worksheet #2 1. A man walks 40 meters north, then 70 meters east, and then 40 meters south. What is his displacement from the starting point? A. 150 meters east B. 150 meters west C. 70 meters east D. 70 meters west 2. As the angle between two concurrent forces decreases from 180 degrees their resultant A. decreases B. increases C. remains the same 3. An object moves a distance of 10 meters in 5 seconds. The average speed of the object is A. 0.5 m/sec B. 2.0 m/sec C. 40 m/sec D. 50 m/sec 4. Starting from rest, object A falls freely for 2.0 seconds, and object B falls freely for 4.0 seconds. Compared with object A, object B falls A. one-half as far B. twice as far C. three times as far D. four times as far 5. A car accelerates uniformly from rest at 3.2 m/sec-sec. When the car has traveled a distance of 40.0 meters its speed will be A. 8.0 m/sec B. 12.5 m/sec C. 16 m/sec D. 128 m/sec 6. Which quantity has both magnitude and direction? A. distance B. speed C. mass D. velocity 7. An object is displaced 12 meters to the right and then 16 meters upward. The magnitude of the resultant displacement is A. 1.3 meters B. 20 meters C. 28 meters D. 4.0 meters 8. An object travels for 8.00 seconds with an average speed of 160.0 meters per second. The distance traveled by the object is A. 20.0 m B. 200. m C. 1,280 m D. 2,560 m 9. The maximum number of components that a single force may be resolved into is A. one C. 2 B. Unlimited D. None 10. As a body falls freely near the surface of the Earth its acceleration A. decreases B. Increases C. remains the same 11. Which is a vector quantity? A. acceleration due to gravity B. mechanical equivalent of heat C. rest mass of an electron D. speed of an object (OVER) 34 12. Starting from rest, an object rolls freely down an incline that is 10 meters long in 2 seconds. The acceleration of the object is approximately A. 5 m/sec B. 5 m/sec2 C. 10 m/sec D. 10 m/sec2 13. An object, initially at rest, falls freely near the Earth’s surface. How long does it take the object to attain a speed of 98 meters per second? A. 0.1 sec B. 10 sec C. 98 sec D. 960 sec 14. The average speed of a runner in a 400-meter race is 8.0 meters per second. How long did it take the runner to complete the race? A.80 sec B.50 sec C.40 sec D. 32 sec 15. A rock is thrown horizontally from the top of a cliff at 12 meters per second. Approximately how long does it take the rock to fall 45 meters vertically? (Assume negligible air resistance.) A. 1.0 sec B. 5.0 sec C. 3.0 sec D. 8.0 sec 16. Two concurrent forces act at right angles to each other. If one of the forces is 40 newtons and the resultant of the two forces is 50.0 newtons the magnitude of the other force must be A. 10 newtons B. 20 newtons C. 30 newtons D. 40 newtons 17. If two 10.0 newton concurrent forces have a resultant of zero the angle between the forces must be A. 0 degrees B. 45 degrees C. 90 degrees D. 180 degrees 18. Acceleration is a vector quantity that represents the time-rate of change in A. momentum B. velocity C. distance D. energy 19. A moving body must undergo a change of A. velocity C. position B. acceleration D. direction 20. A 10.0 kilogram object, starting from rest, slides down a frictionless incline with a constant acceleration of 2.0 m/s2 for 4.0 seconds. During the 4.0 seconds the object moves a total distance of A. 32m B. 16m C. 8.0m D. 4.0m 21. A car is accelerated at 4.0 m/sec-sec from rest. The car will reach a speed of 28 meters per second at the end of A. 3.5 sec B. 7.0 sec C. 14 sec D. 24 sec 22. If a car increases its speed from 15 meters/second to 30 meters/second in 15 seconds the average acceleration during this time is A. 1.0 m/sec2 B. 15 m/sec2 2 C. 30. m/sec D. 45 m/sec2 (OVER) 35 23. An object starting from rest falls freely for 2.0 seconds. During this time its average velocity is A. 4.9 m/sec B. 6.0 m/sec C. 9.8 m/sec D. 19.6 m/sec 24. Assume that an object has no unbalanced force acting on it. Which statement about the object is true? A. The object may be in motion. B. The object must be slowing down. C. The object must be at rest. D. The object may be speeding up. E. Both A&C 25. A softball is thrown straight up, reaching a maximum height of 20 meters. Neglecting air resistance, what is the ball’s approximate vertical velocity when it hits the ground? A. 10 m/sec B. 20 m/sec C. 15 m/sec D. 40 m/sec 26. Which statement about the movement of an object with zero acceleration is true? A. The object must be at rest. B. The object must be slowing down. C. The object may be speeding up. D. The object may be in motion. E. Both A&D 36 Name ______________________________ Date:_____________ Regents Physics Mr. Morgante Uniform Motion/Acceleration Worksheet Show all calculations Place answer in space provided. Equations: v = d/t vf = v1 + at d = vit +1/2 at2 1. A bicycle rider travels at 15 km/h for 1.5 hours. How far did she travel in this time? 1. . 2. A tennis ball starts from a resting position (vi = 0). The ball rolls down a ramp and accelerates at a constant rate of 7 m/s2 The time to reach the bottom of the ramp was 3 sec. What is the balls speed at the bottom of the ramp? 2.__________ 3. The formula for final velocity is vf = vi + at. a. Solve for a SHOW YOUR STEPS OR NO CREDIT. 3. ________ 4. An airplane traveling at 400 m/s is decelerated uniformly at the rate of -25 m/s2. How long does it take for the plane to reach a final velocity of 0 (zero)? 4._________ a. How many meters will the airplane travel before it comes to a stop? a.________ 5. A softball is dropped from a trestle and falls for 4.5 s, before it is caught by Mullen. How fast was the ball traveling when it hit her glove? 5._________ a. How far did the ball fall when dropped from the trestle? a.__________ 37 Name:________________________ Date:_____________ Regents Physics Mr. Morgante 1-D Motion Worksheet #3 (Show All Work) 1. A car traveling at 25 m/s is brought to rest at a constant rate in 20 sec. By applying the brakes. a. What is its acceleration? b. How far did it travel after the brakes were applied? 2. During a 30 second interval, the speed of a rocket rose steadily from 0.1 km/s to 0.5 km/s. How far did the rocket go during this time? 3. A bottle rocket leaves the ground moving straight upward with a velocity of 25 m/s. Ignore air resistance. a. What is the velocity of the rocket after 2.5 seconds? b. How long does it take for the rocket to reach the highest point of its flight? c. What is the total time of flight for the rocket? d. What is the displacement of the rocket from the ground at t=4 seconds? 38 Name________________________________ Regents Physics Kinematics Practice Date________ Mr. Morgante 1. Students in a physics lab collected the following data from a ball starting at rest rolling with uniform acceleration down a 4 m inclined plane (Ignore friction) a. Complete the sketch: +y +x ti = _____ vi = _____ di = _____ tf = _____ vf = _____ df = 4m DATA: Displacement: 0 m 0.75m 1.7m 3m 4m Time: 1.0 secs 1.5 secs 2.0 secs 2.3 secs 0 secs b. What is the velocity of the ball at t = 2.00 secs? Magnitude________ Direction_______ c. What is the displacement of the ball at t = 0.5 secs? Magnitude________ Direction_______ d. What is the acceleration of the ball? Magnitude________ Direction_______ e. Graph the displacement, velocity and acceleration of the ball versus time: (OVER) 39 2. A tennis ball leaves the ground (y = 0) at vi = 30 m/s in the +y direction. 2 Note: Magnitude of g = 10 m/s Sketch a. Calculate the velocity of the ball at t = 4.0 secs Magnitude________ Direction_______ b. How long does it take the ball to reach the top of its arc? b.__________ 3. A race car accelerates uniformly from +40 m/s to +25 m/s in 30 seconds. +x ti = 0 vi = _____ di = _____ tf = _____ vf = _____ df = _____ a. Calculate the total displacement of the car in this 30 –second interval: Magnitude________ Direction_______ b. Calculate the acceleration of the car in this 30 –second interval: Magnitude________ Direction_______ c. Graph the displacement, in the 30-second interval. velocity and acceleration of the car versus time 40 Name:_____________________ Regents Physics Date:_______ Mr. Morgante Ticker Tape Worksheet For #’s 1 - 3.: These questions refer to the tapes shown below. The tapes start on the right hand side of the page. You may use one tape for more than one answer. Explain Your Answer!!!!! A) * B) * C) * * * * * * * * * * * * * * * 1. Which tape shows constant, non-zero acceleration? 2. Which tape shows constant speed? 3. Which tape shows the greatest magnitude of acceleration? 4. What is happening in C) Z:\Physics\Regents Physics\Class Material\Unit 2B 1D Motion 1-7-10.doc 41 * START * * START * START