Physics 2 LS PDF

Topic 2 :Describing Motion: Kinematics in One Dimension Essential idea: Motion may be described and analysed by the use of graphs and equations. Nature of science: Observations: The ideas of motion are fundamental to many areas of physics, providing a link to the consideration of force...

Topic 2 :Describing Motion: Kinematics in One Dimension Essential idea: Motion may be described and analysed by the use of graphs and equations. Nature of science: Observations: The ideas of motion are fundamental to many areas of physics, providing a link to the consideration of forces and their implication. The kinematic equations for uniform acceleration were developed through careful observations of the natural world. Topic 2 :Describing Motion: Kinematics in One Dimension Understandings: Reference Frames and Displacement Average Velocity Instantaneous Velocity Acceleration Motion at Constant Acceleration Solving Problems Freely Falling Objects Graphical Analysis of Linear Motion Topic 2 :Describing Motion: Kinematics in One Dimension Distance and displacement Kinematics is the sub-branch of mechanics which studies only a body's motion without regard to causes. Dynamics is the sub-branch of mechanics which studies the forces which cause a body's motion. Mechanics is the branch of physics which concerns itself with forces, and how they affect a body's motion. The two pillars of mechanics Galileo Newton Kinematics Dynamics (Calculus) 2.1 – Reference Frames and Displacement **Any measurement of position, distance, or speed must be made with respect to a reference frame. 2.1 – Reference Frames and Displacement We make a distinction between distance and displacement. Displacement (blue line) is how far the object is from its starting point, regardless of how it got there. Distance traveled (dashed line) is measured along the actual path. 2.1 – Reference Frames and Displacement Distance and displacement Kinematics is the study of displacement, velocity and acceleration, or in short, a study of motion. A study of motion begins with position and change in position. Consider Freddie the Fly, and his quest for food: d=6m The distance Freddie travels is simply how far he has flown, without regard to direction. Freddie's distance is 6 meters. 2.1 – Reference Frames and Displacement Distance and displacement Distance is simply how far something has traveled without regard to direction. Freddy has gone 6 m. Displacement, on the other hand, is not only distance traveled, but also direction. Distance = 6 m Displacement = 6 m in the positive x-direction This makes displacement a vector. It has a magnitude (6 m) and a direction (+ x-direction). We say Freddie travels through a displacement of 6 m in the positive x-direction. 2.1 – Reference Frames and Displacement Distance and displacement Let’s revisit some previous examples of a ball moving through some displacements… Displacement A x(m) Displacement B x(m) Displacement A is just 15 m to the right (or +15 m for short). Vector Displacement B is just 20 m to the left (or -20 m for short). FYI Scalar Distance A is 15 m, and Distance B is 20 m. There is no regard for direction in distance. 2.1 – Reference Frames and Displacement Distance and displacement Now for some detailed analysis of these two motions… Displacement A x(m) Displacement B x(m) Displacement ∆x (or s) has the following formulas: ∆x = x2 – x1 displacement s = x2 – x1 where x2 is the final position and x1 is the initial position FYI Many textbooks use ∆x for displacement, and some uses s. And don’t confuse s with seconds! 2.1 – Reference Frames and Displacement Distance and displacement ∆x = x2 – x1 displacement where x2 is the final position and x1 is the initial position EXAMPLE: Use the displacement formula to find each displacement. Note that the x = 0 coordinate has been placed on the number lines. 1 2 Displacement A x(m) 2 0 1 Displacement B x(m) SOLUTION: FYI The correct direction (sign) is automatic! 2.2 Average velocity Speed: how far an object travels in a given time interval. Velocity includes directional information: 2.2 Average velocity Speed and velocity Velocity v is a measure of how fast an object moves through a displacement. Thus, velocity is displacement divided by time, and is measured in meters per second (m s-1). v = ∆x / ∆t velocity EXAMPLE: Find the velocity of the second ball (Ball B) if it takes 4 seconds to complete its displacement. SOLUTION: 2.2 Average velocity Speed and velocity From the previous example we calculated the velocity of the ball to be -5 m s-1. Thus, the ball is moving 5 m s-1 to the left. With disregard to the direction, we can say that the ball’s speed is 5 m s-1. We define speed as distance divided by time, with disregard to direction. PRACTICE: A runner travels 64.5 meters in the negative x-direction in 31.75 seconds. Find her velocity, and her speed. SOLUTION: 2.2 Average velocity_HW_pg 24 2.2 Average velocity_HW_pg 25 2.3 Instantaneous Velocity Determining instantaneous and average values for velocity, speed and acceleration Consider a car whose position is changing. A patrol officer is checking its speed with a radar gun as shown. The radar gun measures the position of the car during each successive snapshot, shown in yellow. How can you tell that the car is speeding up? What are you assuming about the radar gun time? 2.3 Instantaneous Velocity Determining instantaneous and average values for velocity, speed and acceleration We can label each position with an x and the time interval between each x with a ∆t. Then vA = (x2 - x1)/∆t, vB = (x3 - x2)/∆t, and finally vC = (x4 - x3)/∆t. Focus on the interval from x2 to x3. Note that the speed changed from x2 to x3, and so vB is NOT really the speed for that whole interval. We say the vB is an average speed (as are vA and vC). vA vB vC ∆t ∆t ∆t x1 x2 x3 x4 2.3 Instantaneous Velocity Determining instantaneous and average values for velocity, speed and acceleration If we increase the sample rate of the radar gun (make the ∆t smaller) the positions will get closer together. Thus the velocity calculation is more exact. We call the limit as ∆t approaches zero in the equation v = ∆x / ∆t the instantaneous velocity. For this level of physics we will just be content with the average velocity. Limits are beyond the scope of this course. You can use the Wiki extensions to explore limits, and derivatives, if interested. 2.3 Instantaneous Velocity The instantaneous velocity →average velocity over an infinitesimally short time interval. These graphs show (a) constant velocity and (b) varying velocity. 2.4 – Acceleration Acceleration Acceleration is the change in velocity over time. (rate of change of velocity. ) a = ∆v / ∆t acceleration a = (v – u) / t where v is the final velocity and u is the initial velocity Since u and v are measured in m/s and since t is measured in s, a is measured in m/s2, or m s-2. FYI Many textbooks use ∆v = vf - vi (OR v2-v1) for change in velocity, vf for final velocity and vi initial velocity. We gets away from the subscripting mess by choosing v for final velocity and u for initial velocity. 2.4 – Acceleration Acceleration a = ∆v / ∆t acceleration a = (v – u) / t where v is the final velocity and u is the initial velocity EXAMPLE: A driver sees his speed is 5.0 m s-1. He then simultaneously accelerates and starts a stopwatch. At the end of 10. s he observes his speed to be 35 m s-1. What is his acceleration? SOLUTION: 2.4 – Acceleration Acceleration a = ∆v / ∆t acceleration a = (v – u) / t where v is the final velocity and u is the initial velocity PRACTICE: (a) Why is velocity a vector? (b) Why is acceleration a vector? SOLUTION:  2.4 – Acceleration (Instantaneous) Determining instantaneous and average values for velocity, speed and acceleration By the same reasoning, if ∆t gets smaller in the acceleration equation, our acceleration calculation becomes more precise. We call the limit as ∆t approaches zero of the equation a = ∆v / ∆t the instantaneous acceleration. For this level of physics we will be content with the average acceleration. See the Wiki for extensions if you are interested! 2.4 – Acceleration Solving problems using equations of motion for uniform acceleration Back in the 1950s, military aeronautical engineers thought that humans could not withstand much of an acceleration, and therefore put little effort into pilot safety belts and ejection seats. An Air Force physician by the name of Colonel Stapp, however, thought humans could withstand higher accelerations. He designed a rocket sled to accelerate at up to 40g (at which acceleration you would feel like you weighed 40 times your normal weight!). 2.4 – Acceleration Solving problems using equations of motion for uniform acceleration The human to be tested would be Stapp himself. An accelerometer and a video camera were attached to the sled. Here are the results: 2.4 – Acceleration Solving problems using equations of motion for uniform acceleration Here are the data. In 1954, America's original Rocketman, Col. John Paul Stapp, attained a then-world record land speed of 632 mph, going from a standstill to a speed faster than a.45 bullet in 5.0 seconds on an especially-designed rocket sled, and then screeched to a dead stop in 1.4 seconds, sustaining more than 40g's of force, all in the interest of safety. There are TWO accelerations in this problem: (a) He speeds up from 0 to 632 mph in 5.0 s. (b) He slows down from 632 mph to 0 in 1.4 s. 2.4 – Acceleration Solving problems using equations of motion for uniform acceleration There are TWO accelerations in this problem: (a) He speeds up from 0 to 632 mph in 5.0 s. (b) He slows down from 632 mph to 0 in 1.4 s. EXAMPLE: Convert 632 mph to m/s. SOLUTION: Use “well-chosen” ones… 1 mile=5280 ft 1 m = 3.28 ft 2.4 – Acceleration Solving problems using equations of motion for uniform acceleration There are TWO accelerations in this problem: (a) He speeds up from 0 to 632 mph in 5.0 s. (b) He slows down from 632 mph to 0 in 1.4 s. EXAMPLE: Find Stapp’s acceleration during the speeding up phase. SOLUTION: EXAMPLE: Find Stapp’s acceleration during the slowing down phase. (deceleration) 2.4 – Acceleration_HW_pg27 2.5 – Motion at Constant Acceleration Equations of motion for uniform acceleration!!! The equations for uniformly accelerated motion are also known as the kinematic equations. They are listed here s = ut + (1/2)at 2 Displacement v = u + at Velocity v2 = u2 + 2as Timeless s = (u + v)t / 2 Average displacement They can only be used if the acceleration a is CONSTANT (uniform). They are used so commonly throughout the physics course that we will name them. 2.5 – Motion at Constant Acceleration Equations of motion for uniform acceleration From a = (v – u)/t we get at = v – u. Rearrangement leads to v = u + at, the velocity equation. Now, if it is the case that the acceleration is constant, then the average velocity can be found by taking the sum of the initial and final velocities and dividing by 2 (just like test grades). Thus average velocity = (u + v) / 2. But the displacement is the average velocity times the time, so that s = (u + v)t / 2, the average displacement equation. 2.5 – Motion at Constant Acceleration Equations of motion for uniform acceleration We have derived v = u + at and s = (u + v)t / 2. Let’s tackle the first of the two harder ones. s = (u + v)t / 2 Given s = (u + u + at)t / 2 v = u + at s = (2u + at)t / 2 Like terms s = 2ut/2 + at 2/ 2 Distribute t/2 s = ut + (1/2)at 2 Cancel 2 which is the displacement equation. Since the equation s = (u + v)t/2 only works if the acceleration is constant, s = ut + (1/2)at 2 also works only if the acceleration is constant. 2.5 – Motion at Constant Acceleration Equations of motion for uniform acceleration We now have derived v = u + at, s = (u + v)t / 2 and s = ut + (1/2)at 2. Let’s tackle the timeless equation. From v = u + at we can isolate the t. v – u = at t = (v – u)/a From s = (u + v)t / 2 we get: 2s = (u + v)t Multiply by 2 2s = (u + v)(v – u) / a t = (v - u)/a 2as = (u + v)(v – u) Multiply by a 2as = uv – u2 + v2 – vu v2 = u2 + 2as 2.5 – Motion at Constant Acceleration Equations of motion for uniform acceleration Just in case you haven’t written these down, here they are again. s = ut + (1/2)at2 Displacement kinematic v = u + at Velocity equations v2 = u2 + 2as Timeless a is constant s = (u + v)t/2 Average displacement We will practice using these equations soon. They are extremely important. 2.5 – Motion at Constant Acceleration_HW_pg 29  2-6 Solving Problems (SOP!!) 1. Read the whole problem and make sure you understand it. Then read it again. 2. Decide on the objects under study and what the time interval is. 3. Draw a diagram and choose coordinate axes. 4. Write down the known (given) quantities, and then the unknown ones that you need to find. 5. What physics applies here? Plan an approach to a solution. 6. Which equations relate the known and unknown quantities? Are they valid in this situation? Solve algebraically for the unknown quantities, and check that your result is sensible (correct dimensions). 7. Calculate the solution and round it to the appropriate number of significant figures. 8. Look at the result—is it reasonable? Does it agree with a rough estimate? 9. Check the units again. 2-6 Solving Problems 2-6 Stopping distance **Estimate the minimum stopping distance for a car, which is important for traffic safety and traffic design. The problem is best dealt with in two parts, two separate time intervals. Thinking distance/Reaction time: The first-time interval Braking distance: →when the driver decides to hit the The second time interval brakes, and ends when the foot →actual braking period when the touches the brake pedal. This is the vehicle slows down (a ≠ 0) "reaction time" during which the speed and comes to a stop. is constant, so a = 0. 2-6 Solving Problems_Stopping distance_pg 32 Reaction time of the driver Stopping distance The initial speed of the car (the final speed is zero) depends The deceleration of the car. For a dry road and good tires, good brakes can decelerate a car at a rate of about 5 m/s2 to 8 m/s2. Calculate the total stopping distance for an initial velocity of 50 km/h (= 14 m/s ≈31 mi/h) and assume the acceleration of the car is -6.0 m/s2 (the minus sign appears because the velocity is taken to be in the positive x direction and its magnitude is decreasing). Reaction time for normal drivers varies from perhaps 0.3 s to about 1.0 s; take it to be 0.50 s. 2-6 Solving Problems_Braking distance_pg 32 Distance when the break are applied Final position when the car stopped 2.7 Freely Falling Objects Determining the acceleration of free-fall experimentally Everyone knows that when you drop an object, it picks up speed when it falls. Galileo did his famous freefall experiments on the tower of Pisa long ago, and determined that all objects fall at the same acceleration in the absence of air resistance. Thus, as the next slide will show, an apple and a feather will fall side by side! 2.7 Freely Falling Objects Determining the acceleration of free-fall experimentally Consider the multi-flash image of an apple and a feather falling in a partial vacuum: If we choose a convenient spot on the apple, and mark its position, we get a series of marks like so: 2.7 Freely Falling Objects Determining the acceleration of free-fall experimentally Now we SCALE our data. Given that the apple is 8 cm 0 cm in horizontal diameter we can superimpose this scale -9 cm on our photograph. Then we can estimate the -22 cm position in cm of each image. -37 cm -55 cm 2.7 Freely Falling Objects Determining the acceleration of free-fall experimentally Suppose we know that the time between images is 0 cm 0.056 s. -9 cm We make a table starting with the raw data columns of t and y. -22 cm t(s) y(cm) t y v We then make.000 0 calculations columns in t,.056 FYI: -9you.056 To find t -37-9 cm TWO need to subtract -161 y and v. t's..112 FYI: Therefore Tofind To -22 findyvtthe youfirst you.056 needentry need to for t is -13 todivide y TWO subtract -232 by BLANK. t. y's. t's. By Byconvention, By convention, convention, CURRENT CURRENT CURRENT y tyMINUS MINUS.168 -37.056 -15 -268 DIVIDED FYI: SameBY PREVIOUS t.CURRENT thing y. t. y. for the first.224 -55.056 -18 -55 cm -321 FYI: Since v = y / t, the first v entry is also BLANK. 2.7 Freely Falling Objects Determining the t(s) y(cm) t y v acceleration of.000 0 free-fall.056 -9.056 -9 -161 experimentally.112 -22.056 -13 -232 Now we plot v.168 -37.056 -15 -268 v vs. t on a.224 -55.056 -18 -321 graph. TIME / sec VELOCITY / cm sec-1.000.056.112.168.224 0 t -50 -100 -150 -200 -250 -300 2.7 Freely Falling Objects Determining the FYI acceleration of The graph v vs. t is linear. Thus a is free-fall constant. experimentally The y-intercept (the initial velocity of the apple) is not zero. But this just v means we don’t have all of the images TIME of the apple. (sec).000.056.112.168.224 VELOCITY (cm/sec) 0 t/s -50 -100 -150 -200 -250 -300 2.7 Freely Falling Objects Determining the FYI acceleration of Finally, the acceleration is the slope free-fall of the v vs. t graph: experimentally a = v = -220 cm/s = -982 cm/s2 v t 0.224 s TIME (sec).000.056.112.168.224 VELOCITY (cm/sec) 0 t/s t = 0.224 s v = -220 cm/s -50 -100 -150 -200 -250 -300 2.7 Freely Falling Objects Determining the acceleration of free-fall experimentally Since this acceleration due to gravity is so important we give it the name g. ALL objects accelerate at -g , where g = 980 cm s-2 in the absence of air resistance. We can list the values for g in three ways: g = 980 cm s-2 magnitude of the freefall g = 9.80 m s-2 acceleration g = 32 ft s-2 g = 9.8 m s-2 2.7 Freely Falling Objects Solving problems using equations of motion for uniform acceleration -General: s = ut + (1/2)at2, and v = u + at, and v2 = u2 + 2as, and s = (u + v)t / 2; -Freefall: Substitute ‘-g’ for ‘a’ in all of the above equations. FYI The kinematic equations will be used throughout the year. We must master them NOW! 2.7 Freely Falling Objects Solving problems using equations of motion for uniform acceleration EXAMPLE: How far will Pinky and the Brain go in 30.0 seconds if their acceleration is 20.0 m s -2? KNOWN FORMULAS a = 20 m/s2 Given s = ut + 12at2 t = 30 s Given v = u + at u = 0 m/s Implicit v2 = u2 + 2as WANTED s=? SOLUTION t is known - drop the timeless eq’n. Since v is not wanted, drop the velocity eq'n: 2.7 Freely Falling Objects Solving problems using equations of motion for uniform acceleration EXAMPLE: How fast will Pinky and the Brain be going at this instant? KNOWN FORMULAS a = 20 m/s2 Given s = ut + 12at2 t = 30 s Given v = u + at u = 0 m/s Implicit v2 = u2 + 2as WANTED v=? SOLUTION t is known - drop the timeless eq’n. Since v is wanted, drop the displacement eq'n: 2.7 Freely Falling Objects Solving problems using equations of motion for uniform acceleration EXAMPLE: How fast will Pinky and the Brain be going when they have traveled a total of 18000 m? KNOWN FORMULAS a = 20 m/s2 Given s = ut + 12at2 s = 18000 m Given v = u + at u = 0 m/s Implicit v2 = u2 + 2as WANTED v=? SOLUTION Since t is not known - drop the two eq’ns which have time in them. 2.7 Freely Falling Objects Solving problems using equations of motion for uniform acceleration EXAMPLE: A ball is dropped off of the Empire State Building (381 m tall). How fast is it going when it hits ground? KNOWN FORMULAS 1 2 2 a = -10 m/s Implicit s = ut + 2at s = -381 m Given v = u + at u = 0 m/s Implicit v2 = u2 + 2as WANTED v=? SOLUTION Since t is not known - drop the two eq’ns which have time in them. 2.7 Freely Falling Objects Solving problems using equations of motion for uniform acceleration EXAMPLE: A ball is dropped off of the Empire State Building (381 m tall). How long does it take to reach the ground? KNOWN FORMULAS a = -10 m/s2 Implicit s = ut + 12at2 s = -381 m Given v = u + at u = 0 m/s Implicit v2 = u2 + 2as WANTED t=? SOLUTION Since t is desired and we have s drop the last two eq’ns. 2.7 Freely Falling Objects Solving problems using equations of motion for uniform acceleration EXAMPLE: A cheer leader is thrown up with an initial speed of 7 m s-1. How high does she go? KNOWN FORMULAS a = -10 m/s2 Implicit s = ut + 12at2 u = 7 m s-1 Given v = u + at v = 0 m/s Implicit v2 = u2 + 2as WANTED s=? SOLUTION Since t is not known - drop the two eq’ns which have time in them. 2.7 Freely Falling Object's Solving problems using equations of motion for uniform acceleration EXAMPLE: A ball is thrown upward at 50 m s-1 from the top of the 300-m Millau Viaduct, the highest bridge in the world. How fast does it hit ground? KNOWN FORMULAS a = -10 m/s2 Implicit s = ut + 12at2 u = 50 m s-1 Given v = u + at s = -300 m Implicit v2 = u2 + 2as WANTED v=? SOLUTION Since t is not known - drop the two eq’ns which have time in them. 2.7 Freely Falling Objects Solving problems using equations of motion for uniform acceleration EXAMPLE: A ball is thrown upward at 50 m s-1 from the top of the 300-m Millau Viaduct, the highest bridge in the world. How long is it in flight? KNOWN FORMULAS 1 2 a = -10 m/s 2 Implicit s = ut + 2at u = 50 m s-1 Given v = u + at v = -92 m s-1 Calculated v2 = u2 + 2as WANTED t=? SOLUTION Use the simplest t equation. 2.7 Freely Falling Objects_HW_pg 34 2.7 Freely Falling Objects_HW_pg 35 2.7 Freely Falling Objects_exp1 2.7 Freely Falling Objects_exp 2 2.7 Freely Falling Objects_exp 3 2.7 Freely Falling Objects_exp 4 Free falls When stuff feels ONLY gravity!! 2.7 Graphical Analysis of Linear motion Sketching and interpreting motion graphs The slope of a displacement-time graph is the velocity. The slope of the velocity-time graph is the acceleration. We already did this example with the falling feather/apple presentation. You will have ample opportunity to find the slopes of distance-time, displacement-time and velocity-time graphs in your labs.  2.7 Graphical Analysis of Linear motion Sketching and interpreting motion graphs EXAMPLE: Suppose Freddie the Fly begins at x = 0 m, and travels at a constant velocity for 6 seconds as shown. Find two points, sketch a displacement vs. time graph, and then find and interpret the slope and the area of your graph. t = 0, x = 0 x/m t = 6 s, x = 18 SOLUTION: The two points are (0 s, 0 m) and (6 s, 18 m). The sketch is on the next slide. 2.7 Graphical Analysis of Linear motion Sketching and interpreting motion graphs SOLUTION: 27 24 21 18 x/m 15 Rise 12 9 s = 18 - 0 6 t=6-0 s = 18 m 3 Run t = 6 s 0 0 1 2 3 4 5 6 7 8 9 t/s 2.7 Graphical Analysis of Linear motion Sketching and interpreting motion graphs The area under a velocity-time graph is the displacement. The area under an acceleration-time graph is the change in velocity. Graph Slope Area under the graph x vs t Velocity - v vs t Acceleration Displacement a vs t - Change in velocity 2.7 Graphical Analysis of Linear motion Sketching and interpreting motion graphs EXAMPLE: Calculate and interpret the area under the given v vs. t graph. Find and interpret the slope. VELOCITY (ms-1 ) 50 SOLUTION: 40 30 20 10 0 t 0 5 10 15 20 TIME (sec)  Summary of Topic 2 Kinematics is the description of how objects move with respect to a defined reference frame. Displacement is the change in position of an object. Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time.