Chapter : Motion in One Dimension Review: velocity can either be constant or changing. What is the mathematical meaning of v avg? The equation of a straight line is y = mx + b. From the definition of average speed, we find that x solve for x v = x = x x t t = vt x = vt + x = Plotting x(t) vs. t Since pictures are worth a words, plotting displacement vs. time leads to a deeper understanding of motion and is indispensible. I cannot stress enough just how important it is to physically see and understand plots. Plots/graphs are a powerful technique in showing tends in the overall behavior of a system. Let s take some steps to mastering these concepts. Case : Stationary cart A slope of zero implies that the speed is zero. From the equation of the line, x(t) = vt + x = x for all time Case : Cars moving at constant velocity Suppose there are two cars on the highway with one traveling faster than the other; the faster car (v ) passes the slower one (v ). Graphically, the plots are lines with constant slopes where the faster car (shown in blue) has the higher speed v and therefore, the steeper slope when compared to the slower (v ) car (shown in red). x (t) = v t + x and (t) = t + x As one already knows, a faster car travels farther distances (v t) than a slower one (v t). Key points The slope tells us how fast an object is moving. x steeper slope faster speed; less steep slope slower speed The vertical change of a linear line ( x = vt) tells us the change in the horizontal distance that an object has moved. Higher speeds imply larger distance traveled whereas a lower speed implies a shorter distance ( x = v t > x = v t). If a car starts slowing down or speeding up, the slope of the line changes. Increase in speed, the slope increases; decrease in speed, the slope decreases. Graphically, how do these linear lines change if an object moves in the negative direction? positive slope moving to the right; negative slope moving to the left v Plotting position and velocity vs. time graphs (i) Interpret the following x vs. t graph by writing a very short story of what is happening. (ii) Draw a (i) motion diagram, (ii) rank the speeds, and (iii) plot v vs. t plot. Plot : after driving miles at 6 mph on the interstate, I stopped at a tea shop for black tea. When I got back on the road minutes later, I was slowed to 3 mph by a construction zone for minutes. Finally, I was able to get back up to a cruising speed and I got off my exit 5 miles from home, I reached school. After searching for minutes, I realized that I left my physics homework and lab report at home so I drove back, without stops or construction delays, at 75 mph..
Plot : a football is kickoff, Juan (red line) catches it on the yd line, and then heads up the field at a sprint. Fred (blue line) runs towards Juan in an attempt to tackle him, but misses as Juan crosses the 5 yd line. Fred vainly tries to catch up, but Juan scores. Foolishly, Fred continues to run after Juan has already scored. Example. Sketch x vs. t and v vs. t for the following motions. Include a numerical scale on both axes with units that are reasonable for this motion. Some numerical information is given in the problem, but for other quantities, make reasonable estimates. Note: A sketched graph is hand-drawn, rather than laid out with a ruler. Even so, a sketch must be neat, accurate, and include axis labels. a. A Carmen walks to the bus stop, waits for the bus, and then rides to campus. Assume that all the motion is along a straight street. b. Quarterback Joe drops back yards from the line of scrimmage, and then throws a pass yards to a receiver John, who catches it and sprints yards to the goal. Draw your graph for the football. Think carefully about what the slopes of the lines should be. Acceleration Every time one changes their velocity, acceleration has occurred. Since acceleration is vector, it has both magnitude and direction: change in velocity v magnitude = a acceleration = a = time interval direction = left/right or ± Units: [a] = [v/t] = (m/s)/s We will focus only on changing velocity related to a speed change, not direction change (will do this in lab, not lecture). Magnitude of acceleration One has an intuitive feel for how small or large acceleration is because it depends on how quickly velocity is changed. When speeding up or slowing down, our bodies can measure the magnitude of the acceleration because one feels a force acting on our bodies. When braking occurs while driving a car, we feel our body being forced forwards. By measuring this forward motion, one can determine the strength/magnitude of the acceleration. When applying the brakes in a car in order to change the speed of a car, there are two ways to slow down: slowing down over a (i) long or (ii) short time interval. Physically, applying the brakes over an extended time interval (pushing the brake pedal lightly), our bodies move slightly forward as if there was a small force acting on us. On the other hand, braking over a very short time interval (i.e. slamming on the brakes to avoid a child that has suddenly run out in front of your car), you ve very quickly have changed your velocity and consequently feel a very strong force pushing you forward. A greater change in velocity implies a higher acceleration this is what is meant by the magnitude of the acceleration. Note, if your car moves at a constant velocity one does not feel a force pushing on you. Mathematically, changing the velocity over a long-time interval results in a small acceleration (or deceleration) while as changing it over a short time interval gives a large acceleration:.
v = a a long time interval small acceleration feel a small force acting on you v = short time interval large acceleration feel a large force acting on you Direction of acceleration If we use our bodies as accelerometers, the direction of the acceleration is always determined as being in the opposite direction of your body s direction. Suppose you are driving in a car and it suddenly increase your speed very quickly, your body feels a backwards push. According to our definition, the direction of the acceleration is opposite the direction of your body and therefore, the acceleration and velocity point in the same direction when speeding up. If driving down the street and suddenly have to stop very quickly, your body moves forward and the acceleration is opposite of my body's motion; the acceleration and velocity now point in opposite directions. In summary, we now add acceleration to the motion diagrams and simplify the drawing of them. In everyday language, when an object is speeding up, we usually call this accelerating whereas when an object is slowing down, deceleration. In general, I will not differentiate between acceleration and deceleration since both cause changes in velocity. What does differentiate between speeding up or slowing down is the direction of the acceleration. Therefore, positive acceleration is when the velocity and acceleration point in the same direction whereas negative acceleration is with them in opposite directions. Example Suppose two cars start with the same speed of 3 m/s but decelerate at two different rates of a and a where a > a. What does this mean? The acceleration tells us how quickly the velocity is changed. Suppose further that v 5 m/s m/s a = = and a = s s Picture wise it means that Plotting v(t) vs. t Graphically, acceleration and velocity have similar behaviors. The slope of the graph x vs. t is the velocity, which tells us that the size of the slope is the speed and the sign of the slope (±) the direction of motion. In a similar manner, by starting with the acceleration equation and solving for the velocity, a linear equation pops out and can be interpreted graphically: v a = v(t) = v + at Graphically, what does acceleration mean?. The magnitude of the slope is the acceleration. A steeper slope implies a higher acceleration and therefore, steeper slope higher acceleration; less steep slope lower acceleration The vertical change of the linear line ( v = at) tells us how much the velocity has changed. Higher accelerations imply larger changes in velocity whereas a lower acceleration implies a smaller change in velocity..3
. Speeding up implies a positive acceleration. Both the velocity and acceleration have to point in the same direction. The change in velocity is given by v = at, so the velocity vector increases by the amount of v = +at: Slowing down implies a negative acceleration. The velocity and acceleration are pointing in opposite directions. Equation wise we write v = v at, so the velocity vector decreases by the amount of v = at: 3. The direction of the acceleration vector is given by the slope direction. Positive/negative slope implies positive/negative acceleration. If we now combine all three types of motion plots together, here is what we see in terms of speeding up or slowing down. DEMO Pasco Carts with attached spring Questions a. Can the cart have acceleration in the west direction and simultaneously have a velocity in the east? b. Can the cart be increasing in speed as its acceleration decreases? If so, give an example; if not explain why? c. Can the velocity of an object reverse direction when its acceleration is constant? If so, give an example; if not, explain why. Example. Below are three velocity-versus-time graphs. For each, draw the corresponding accelerationversus-time graph and draw a motion diagram below the graphs..4
One-Dimensional Motion with constant acceleration Our goal is not to derive the kinematic equations for linear motion but to know how they work and what they mean. Using our definitions x, v = x/, a = v/, we will be able to describe motion completely, which implies that we will know x, v, a, and t for the entire motion. The KINEMATIC equations for LINEAR MOTION are v = v + at x x = v + at v v = a x x = (v + v )t These equations are critical for understanding motion and you will be quizzed and tested on your understanding both conceptually and your working knowledge of these. It is important that we get to know these intimately. So, let s organize and digest these equations.. Although there are 5 variables, each of the kinematics equations contains four of them. Goal: if three are known, then the fourth variable can be solved for.. One way to organize this information is to create data tables that look like Data x a v v o t The idea is to insert the know quantities into this data table and insert a question mark? for the unknown variable. 3. From the data table information, one needs to identify which equation to use in order to solve for the unknown variable. Let s summarize the information in a kinematics table. Starting with the first equation v = v + at, we note that this equation contains four variables: v, v o, a, and t. The variable that is missing is the displacement x. If in a particular problem the displacement is not given, one can use this equation to solve for the unknown variable in terms of three known variables. The table indicates this by using to indicate that the variable is present and an X to indicate that it is not present in the equation. In other words, we know which equation to use by a variable s absence in the equation. Kinematics x a v v o t v = v + at X v = v + a x X x = vt + at X x = (v + v )t X.5
4. Reasoning Strategy for Applying the Equations of Kinematics Step : Draw a motion diagram and plot v vs. t of the situation and interpret the question. Step : Define the direction of motion (velocity) to be always positive! Step 3: Create a Data Table with all units converted into the mks-system. Step 4: Use the kinematics table to organize your thinking and select the correct equation to use. Step 5: Solve for the correct variable and interpret your result. Let s look at several examples to get the process of organizing some of these data tables, and picking out the correct equation to use. For this first problem I will go through it completely, and then we will look at two others where I will ask you to apply the Reasoning Strategy. I will typically say Read the question first and then apply the Reason Strategy. Example.3 You're driving down the highway late one night at m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is.5 s, and the maximum deceleration of your car is m/s. a. Plot a motion diagram and a velocity vs. time plot. b. How much distance is between you and the deer when you come to a stop? Solution A visual overview of your car s motion should include a motion diagram and v vs. t plot, and a list of values is shown below. This is a two-part problem because the car first moves at constant velocity and then decelerates. A motion diagram or the plot of v vs. t will explicitly show this. We will first find the car s displacement x during your reaction time when the car s deceleration is zero and the displacement x as you bring the car to rest with maximum deceleration. Finding x : setting up a Data Table from the motion diagram, we write down Since we have all of the variables, we can choose any equation from the kinematics table: x = vt+ at = = m = x Finding x : repeating the above process, we write down x a v v t Now, we see that we do? -m/s m/s X not know t and therefore, this determines which equation we use. Using the kinematics table we immediately choose equation v = v + a x = ( m/s) + ( m/s ) x x = 3 m Find the distance between car and deer: x a v v t? m/s m/s.5s x = x x = 35 3 = 5m deer car Question.4 A light-rail train going from one station to the next on a straight section of track accelerates from rest at. m/s for m. It then proceeds at constant speed for m before slowing down at. m/s until it stops at the station. How long does it take the train to go from one station to the next? Solution.6
There are three parts to this motion. This is easiest to see if we plot the v vs. t graph and/or the motion diagram: a. To solve for the total time between stations, define ttotal = t + t + t3. Time t Setup a data table and use it to determine the correct equation (via kinematics table). x a v v t m.m/s X? Since the dash is with the final velocity v, the kinematics table immediately tells us to use x x x = vt + a t t = t = = = s = t a a. Time t Setup a data table x a v 3 v t m X X? Because there are two X s in the data table, that means we will need to find the initial velocity (note that this velocity is the same as the final velocity of interval-) before finding the time t. Using the data table from interval-, we get x a v v t m.m/s? s Using this information, we immediately solve for v : v = v + a t = (. m/s )( s) = m/s Now we substitute this back into the second data table and get (note v = v3) x a v 3 v t m m/s m/s? Because we have all of the data, we can use any equation and I choose x m x= vt + a t t = = = 5 s = t v m/s Time t Setup a data table and determine the correct equation. x 3 a 3 v 4 v 3 t 3 X -. m/s m/s? which immediately leads to the equation to solve for C: v4 v3 ( m/s) v4= v3+ at 3 3 t3= = = s = t a3. m/s So the total time for the time to go from one station to the next is t + t + t = s + 5s + s = 8 s = t 3 total FREE FALL Aristotle attempted to clarify motion by classification. However, he only 3.7
Hypothesis Predicated never performed a single expt to test his ideas Experiment Galileo came, years later and challenged these ideas because he did do experiments. Aristotle claimed that heavier objects fell faster than lighter ones. DEMO Drop a heavy and light object simultaneously. Many people have heard about the famous experiment where Galileo dropped two objects (lead and wood) of the leaning Tower of Pisa - this clearly untrue. What is certain is that Galileo used frictionless incline planes to prove his ideas. When objects are dropped, they clearly hit the ground due to the force of gravity. The force of gravity effects all objects that have mass. (If time, comment on the difference between Newtonian and GR.) When air resistance is neglected, we call it FREE FALL. A strange thing at gravity is that it is independent of mass and therefore, all objects will hit the ground at the same time when either Two objects are dropped from rest (e.g., my son and a Sumo wrestler) Two objects are thrown down at the same speed. ALL OBJECTS accelerate towards the earth at a constant rate of acceleration a = g = 9.8 m/s If objects are close to the surface of the earth, the acceleration is constant and points towards the center of the earth. So objects that are moving against the acceleration due to gravity, clearly slow down, whereas moving with the acceleration, clearly speed-up. The sign conventions for the acceleration (negative pointing down) and velocity (positive for up and negative for down) are shown in the diagram. Question: suppose a rifle was fired straight downwards from a high-altitude balloon with a muzzle speed of m/s. What is the ACCELERATION of the bullet after sec? What is the speed after 3 s? Because the acceleration due to gravity is a constant, all the previous equations derive for kinematics immediately transfers. Setting a = g, GUARANTEES that acceleration due to gravity always points downward. The kinematic equations for FREE FALL read v = v gt y y = v gt v v = g y y = (v + v )t Check Question A ball is thrown vertically with v y. Plot and interpret y-, v-, a- vs. t. Graph y vs. t shows that the ball at the peak of the curve has zero slope (v peak = ). The trajectory h is clearly quadratic in nature (parabolic) and is completely symmetric with going upwards vs. downwards. That is, distance ( y up = y down) and time (t up = t down) are symmetric, except for direction in vector quantities. Graph v y vs. t clearly shows three things: (i) v peak = at t peak since it intersects the axis at v y =. (ii) As the ball moves up, the speed of the ball decreases whereas after the peak, the ball speeds up. This means that the velocity vector changed directions. (iii) The initial speed of the ball is equal and opposite to the speed of the ball just before it hits the ground (v initial = v final)..8
Example.5 A football is kicked straight up into the air; it hits the ground 5. s later. a. What was the greatest height reachd by the ball? Assume it is kicked from ground level. b. With what speed did it leave the kicker's foot? Solution One of the biggest red flags in all of freefall/projectile motion problems is that the initial speed must be given. Look at the kinematics table and you will see that every equation contains v. If you do not have v, it will require one to set up a complicated equation that involves solving the quadratic equation. At times, there is no other choice; however, one can use physics to solve this problem without having to grind out a complicated equation. Physics is NOT mathematics, so use your physics thinking to solve problems and do not think in terms of equations! a. We are interested in solving for the maximum height. The problem description gives very little information except the total flight time (i.e., the time the ball is in the air), being 5. s. However, there are some pieces that we can add. We know that at maximum height, the ball s speed is zero. Since the trajectory is symmetric, one can look at the ball going up or equally, see the ball coming down. Furthermore, we get two pieces of information from this: (i) the time going up is equal to the time coming down, t up = t down =.6 s. (ii) The speed of the ball initially must also be the speed of the ball coming down (just before hitting the ground). Armed with these pieces of the puzzles, I will draw the following picture along with the corresponding data tables to organize our thinking for solving for the maximum height. Data table for upward motion y a v v t? -g X.6 s Data table for downward motion What do we know: Upward distance must equal the downward distance but opposite in sign: y = y The initial velocity of the ball must be equal and opposite the final speed of the ball just before hitting the ground: v = v 3 Note that in the upwards motion data table, the initial velocity v is not given. As stated earlier, this is a major algebraic thinking problem without the initial speed. Other the other hand, look at the downwards motion data table, it has the initial speed v =! So, I go along and solve for the maximum height for the downwards motion. Using our procedure, since the X is with the final velocity v 3, the kinematics table immediately tells us to use y = v t + at = v t gt = (9.8 m / s )(.6s) = 33 m = y The maximum height is 33 m (= + y ), where the minus sign just tells us that the ball went downwards. b. To get the maximum speed, we solve for v 3 (= v ) If we now reverse the problem and think of it as going up only, then the final speed at the peak of the trajectory is zero. Setting up the data table leads to y a v 3 v t -33m -g?.6 s This means that we can use any equation and I pick the easiest equation possible: v = v gt v = gt = (9.8 m / s )(.6s) = 5 m/s = v 3 y a v 3 v t? -g X.6 s 3 3.9