# Motion - DP Physics

Topic 2: Mechanics 2.1 Motion 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: Mechanics 2.1 Motion Understandings: Distance and displacement Speed and velocity Acceleration Graphs describing motion Equations of motion for uniform acceleration Projectile motion Fluid resistance and terminal speed

Topic 2: Mechanics 2.1 Motion Applications and skills: Determining instantaneous and average values for velocity, speed and acceleration Solving problems using equations of motion for uniform acceleration Sketching and interpreting motion graphs Determining the acceleration of free-fall experimentally Analysing projectile motion, including the resolution of vertical and horizontal components of acceleration, velocity and displacement Qualitatively describing the effect of fluid resistance on falling objects or projectiles, including reaching terminal speed Topic 2: Mechanics 2.1 Motion Guidance: Calculations will be restricted to those neglecting air resistance Projectile motion will only involve problems using a

constant value of g close to the surface of the Earth The equation of the path of a projectile will not be required Data booklet reference: v = u + at s = ut + (1/2)at2 v2 = u2 + 2as s = (1/2)(v + u) / t Topic 2: Mechanics 2.1 Motion International-mindedness: International cooperation is needed for tracking shipping, land-based transport, aircraft and objects in space Theory of knowledge: The independence of horizontal and vertical motion in projectile motion seems to be counter-intuitive. How do scientists work around their intuitions? How do scientists make use of their intuitions? Topic 2: Mechanics

2.1 Motion Utilization: Diving, parachuting and similar activities where fluid resistance affects motion The accurate use of ballistics requires careful analysis Biomechanics (see Sports, exercise and health science SL sub-topic 4.3) Quadratic functions (see Mathematics HL sub-topic 2.6; Mathematics SL sub-topic 2.4; Mathematical studies SL sub-topic 6.3) The kinematic equations are treated in calculus form in Mathematics HL sub-topic 6.6 and Mathematics SL sub-topic 6.6 Topic 2: Mechanics 2.1 Motion Aims: Aim 2: much of the development of classical physics has been built on the advances in kinematics Aim 6: experiments, including use of data logging, could include (but are not limited to): determination of g, estimating speed using travel timetables,

analysing projectile motion, and investigating motion through a fluid Aim 7: technology has allowed for more accurate and precise measurements of motion, including video analysis of real-life projectiles and modelling/simulations of terminal velocity Topic 2: Mechanics 2.1 Motion Distance and displacement Mechanics is the branch of physics which concerns itself with forces, and how they affect a body's motion. 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. The two pillars of mechanics Galileo Kinematics Newton

Dynamics (Calculus) Topic 2: Mechanics 2.1 Motion 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: t a l o oc h c d e lt e

M hip c e d=6m The distance Freddie travels is simply how far he has flown, without regard to direction. Freddie's distance is 6 meters. Topic 2: Mechanics 2.1 Motion 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. Topic 2: Mechanics 2.1 Motion Distance and displacement Lets 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.

Topic 2: Mechanics 2.1 Motion 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 s = x2 x1 displacement where x2 is the final position and x1 is the initial position FYI Many textbooks use x for displacement, and IB uses s. Dont confuse the change in with the uncertainty symbol. And dont confuse s with seconds!

Topic 2: Mechanics 2.1 Motion Distance and displacement displacement where x2 is the final position and x1 is the initial position x = x2 x1 s = x2 x1 EXAMPLE: Use the displacement formula to find each displacement. Note that the x = 0 coordinate has been 2 placed on the number lines. 1 Displacement A 2 Displacement B 0 x(m)

1 SOLUTION: For A: s = (+10) (-5) = +15 m. FYI For B: s = (-10) (+10) = -20 m. The correct direction (sign) is automatic! x(m) Topic 2: Mechanics 2.1 Motion 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). velocity v = x / t v=s/t EXAMPLE: Find the velocity of the second ball (Ball B) if it takes 4 seconds to complete its displacement.

SOLUTION: For B: s = (-10) (+10) = -20 m. But t = 4 s. Therefore v = -20 m / 4 s = -5 m s-1. Note that v inherits its direction from s. Topic 2: Mechanics 2.1 Motion 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 balls 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: Her velocity is -64.5 / 31.75 = - 2.03 m s-1. Her speed is 64.5 / 31.75 = 2.03 m s-1. Topic 2: Mechanics

2.1 Motion Acceleration Acceleration is the change in velocity over time. a = v / t a = (v u) / t acceleration 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 in IB format, a is measured in m s-2. FYI Many textbooks use v = vf - vi for change in velocity, vf for final velocity and vi initial velocity. IB gets away from the subscripting mess by choosing v for final Topic 2: Mechanics 2.1 Motion Acceleration a = v / t

a = (v u) / t acceleration 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: Label each number with a letter: v = 35 m s-1, u = 5.0 m s-1, and t = 10. s. Next, choose the formula: a = (v u) / t. Now substitute and calculate: -2 Topic 2: Mechanics 2.1 Motion Acceleration a = v / t a = (v u) / t

acceleration 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: (c) Velocity is a displacement over time. Since displacement is a vector, so is velocity. (d) Acceleration is a change in velocity over time. Since velocity is a vector, so is acceleration. Topic 2: Mechanics 2.1 Motion 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!). Topic 2: Mechanics 2.1 Motion 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: Topic 2: Mechanics 2.1 Motion 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. Topic 2: Mechanics 2.1 Motion 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 632 mi 5280 ft 1 m 1 h = 280 m 3.28 ft 3600. s s 1h

1 mi EXAMPLE: Was Stapp more uncomfortable while he was speeding up, or while he was slowing down? SOLUTION: While slowing down. Why? Topic 2: Mechanics 2.1 Motion 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 Stapps acceleration during the speeding up phase. SOLUTION: v = v f - v i = 280 m/s - 0 m/s = 60 m/s2 a= t 5s t EXAMPLE: Find Stapps acceleration during the slowing

down phase. a = v - u = 0 m/s - 280 m/s = - 200 m s-2 1.4 s t Topic 2: Mechanics 2.1 Motion 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? Topic 2: Mechanics 2.1 Motion 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 = (x4 - x3)/t. vC 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 (as vspeed vA vC are vA and vC). B t x1 x2 t x3 t

x4 Topic 2: Mechanics 2.1 Motion 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. Topic 2: Mechanics 2.1 Motion 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! Topic 2: Mechanics 2.1 Motion 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. Topic 2: Mechanics 2.1 Motion 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. Topic 2: Mechanics 2.1 Motion

Equations of motion for uniform acceleration We have derived v = u + at and s = (u + v)t / 2. Lets 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. Topic 2: Mechanics 2.1 Motion 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. Lets 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 FOIL v2 = u2 + 2as Cancel (uv = vu) Topic 2: Mechanics 2.1 Motion Equations of motion for uniform acceleration Just in case you havent written these down, here they are again. s = ut + (1/2)at2

v = u + at v2 = u2 + 2as s = (u + v)t/2 kinematic Displacement equations Velocity a is constant Timeless Average displacement We will practice using these equations soon. They are extremely important. Before we do, though, we want to talk about freefall and its special acceleration g. Topic 2: Mechanics 2.1 Motion 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! Topic 2: Mechanics 2.1 Motion Determining the acceleration of free-fall experimentally Consider the multiflash 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:

Topic 2: Mechanics 2.1 Motion Determining the acceleration of free-fall experimentally Now we SCALE our data. Given that the apple is 8 cm in horizontal diameter we can superimpose this scale on our photograph. Then we can estimate the position in cm of each image. 0 cm -9 cm -22 cm -37 cm -55 cm

Topic 2: Mechanics 2.1 Motion Determining the acceleration of free-fall experimentally Suppose we know that the time between images is 0.056 s. We make a table starting with the raw data columns of t and y. We then make calculations columns in t, y and v. 0 cm -9 cm t(s) y(cm) .000

t -22 cm y v 0 -37-9 cm TWO -161 .056 -9you .056 FYI: To find t need to subtract t's. Therefore first entry for

t is .112 -22 .056 -13 -232 FYI: Tofind find vtthe you need to y TWO by To y you need todivide subtract BLANK.

t. By CURRENT y y's. Byconvention, convention, CURRENT MINUS t's. By convention, CURRENT tyMINUS .168 -37 .056 -15 -268 DIVIDED t. y. FYI:

SameBY thing for the first PREVIOUS y. t.CURRENT .224 -55 .056 -18 -321 -55 cm FYI: Since v = y / t, the first v entry is also BLANK. Topic 2: Mechanics 2.1 Motion VELOCITY / cm sec-1 Determining the

acceleration of free-fall experimentally Now we plot v v vs. t on a graph. 0 -50 -100 -150 -200 -250 -300 .000 t(s) y(cm) t y

v .000 0 .056 -9 .056 -9 -161 .112 -22 .056

-13 -232 .168 -37 .056 -15 -268 .224 -55 .056 -18

-321 .056 TIME / sec .112 .168 .224 t Topic 2: Mechanics 2.1 Motion Determining the acceleration of free-fall experimentally VELOCITY (cm/sec)

v 0 -50 -100 -150 -200 -250 -300 .000 FYI The graph v vs. t is linear. Thus a is constant. The y-intercept (the initial velocity of the apple) is not zero. But this just means we dont have all of the images TIME of the apple.

(sec) .056 .112 .168 .224 t/s Topic 2: Mechanics 2.1 Motion VELOCITY (cm/sec) v 0 -50 -100 -150

-200 -250 -300 .000 FYI Finally, the acceleration is the slope of the v vs. t graph: a = v = -220 cm/s = -982 cm/s2 0.224 s t .056 TIME (sec) .112 t = 0.224 s .168 .224

t/s v = -220 cm/s Determining the acceleration of free-fall experimentally Topic 2: Mechanics 2.1 Motion 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 g = 9.80 m s-2 g = 32 ft s-2

We usually round the metric value to 10: g = 10. m s-2 Hammer and feather drop Apollo 15 magnitude of the freefall acceleration Topic 2: Mechanics 2.1 Motion 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! v=u s = ut + + at 1 at 2 2 as 2 + v2 = u 2

Topic 2: Mechanics 2.1 Motion 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 Implicit u = 0 m/s v2 = u2 + 2as s=? WANTED t is known - drop the timeless eqn. Since v is not wanted, drop the velocity eq'n:

SOLUTION s = ut + 12at2 s = 0(30) + 12 20(30)2 s = 9000 m Topic 2: Mechanics 2.1 Motion Solving problems using equations of motion for uniform acceleration EXAMPLE: How fast will Pinky and the Brain be going at this instant? KNOWN a = 20 m/s2 Given Given t = 30 s Implicit u = 0 m/s FORMULAS s = ut + 12at2 v = u + at

v2 = u2 + 2as v=? WANTED t is known - drop the timeless eqn. Since v is wanted, drop the displacement eq'n: SOLUTION v = u + at v = 0 + 20(30) v = 600 m s-1 Topic 2: Mechanics 2.1 Motion 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

s = ut + 12at2 a = 20 m/s2 Given v = u + at s = 18000 m Given v2 = u2 + 2as Implicit u = 0 m/s v=? SOLUTION WANTED Since t is not known v2 = u2 + 2as drop the two eqns which v2 = 02 + 2(20)(18000) have time in them. v = 850 m s-1 Topic 2: Mechanics 2.1 Motion 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 s = ut + a = -10 m/s Implicit 2at v = u + at s = -381 m Given v2 = u2 + 2as Implicit u = 0 m/s v=? SOLUTION WANTED Since t is not v2 = u2 + 2as known - drop the

v2 = 02+ 2(-10)(-381) two eqns which -1 v = -87 m s have time in them. Topic 2: Mechanics 2.1 Motion 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 a = -10 m/s2 Implicit s = -381 m Given Implicit u = 0 m/s

FORMULAS s = ut + 12at2 v = u + at v2 = u2 + 2as t=? WANTED Since t is desired and we have s drop the last two eqns. SOLUTION s = ut + 12at2 -381 = 0t + 12 (-10)t2 t = 8.7 s Topic 2: Mechanics 2.1 Motion 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 s = ut + 12at2 a = -10 m/s2 Implicit v = u + at Given u = 7 m s-1 v2 = u2 + 2as Implicit v = 0 m/s s=? WANTED Since t is not known drop the two eqns which have time in them. SOLUTION v2 = u2 + 2as 02 = 72 + 2(-10)s s = 2.45 m

Topic 2: Mechanics 2.1 Motion 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 s = ut + 12at2 a = -10 m/s2 Implicit v = u + at u = 50 m s-1 Given v2 = u2 + 2as Implicit s = -300 m v=? WANTED Since t is not known drop the two eqns which have time in them. SOLUTION

v2 = u2 + 2as v2 = 502 + 2(-10)(-300) v = -90 m s-1 Topic 2: Mechanics 2.1 Motion 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 2 s = ut + a = -10 m/s Implicit 2at

v = u + at Given u = 50 m s-1 v2 = u2 + 2as v = -90 m s-1 Calculated WANTED Use the simplest t equation. t=? SOLUTION v = u + at -90 = 50 + (-10)t t = 14 s Topic 2: Mechanics 2.1 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. Topic 2: Mechanics 2.1 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. x/m t = 6 s, x = 18 t = 0, x = 0 SOLUTION: The two points are (0 s, 0 m) and (6 s, 18 m). The sketch is on the next slide. Topic 2: Mechanics

2.1 Motion Sketching and interpreting motion graphs x/m SOLUTION: 27 24 21 18 15 12 9 6 3 0 Rise s = 18 - 0 s = 18 m

t=6-0 Run t = 6 s 0 1 2 3 4 5 6 7 t/s The slope is rise over run or 18 m / 6 s Thus the slope is 3 m s-1, which is interpreted as Freddies velocity.

8 9 Topic 2: Mechanics 2.1 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. You will have ample opportunity to draw distance-time, displacement-time and velocity-time graphs in your labs. Topic 2: Mechanics 2.1 Motion Sketching and interpreting motion graphs VELOCITY (ms-1 ) EXAMPLE: Calculate and interpret the area under the

given v vs. t graph. Find and interpret the slope. 50 SOLUTION: 40 The area of a 30 triangle is 20 10 A = (1/2)bh. 0 10 0 5 Thus TIME (sec) A = (1/2)(20 s)(30 m/s) = 300 m. This is the displacement of the object in 20 s. The slope is (30 m/s) / 20 s = 1.5 m s-2. This is the acceleration of the object. 15

20 t Topic 2: Mechanics 2.1 Motion Qualitatively describing the effect of fluid resistance on falling objects or projectiles, including reaching terminal speed -Students should know what is meant by terminal speed. -This is when the drag force exactly balances the weight. Topic 2: Mechanics 2.1 Motion "A female Blue Whale weighing 190 metric tonnes (418,877lb) and measuring 27.6m (90ft 5in) in length suddenly materialized above the

Southern Ocean on 20 March 1947." y Guinness World Records. Falkland Islands Philatelic Bureau. 2 March 2002. At first, v= 0. Qualitatively describing the effect of fluid resistance on falling objects or projectiles, W including reaching terminal speed Then, as v y Suppose a blue whale suddenly increases, so D materializes high above the ground. does D. The drag force D is proportional to

the speed squared. W v Thus, as the whale picks up speed, v reaches a the drag force increases. maximum value, D Once the drag force equals the called terminal y whales weight, the whale will stop speed. accelerating. D = W. It has reached terminal speed. W vterminal Topic 2: Mechanics 2.1 Motion Determine relative velocity in one and two dimensions Suppose you are a passenger in a car on a perfectly

level and straight road, moving at a constant velocity. Your velocity relative to the pavement might be 60 mph. Your velocity relative to the driver of your car is zero. Whereas your velocity relative to an oncoming car might be 120 mph. Your velocity can be measured relative to any reference frame. A Topic 2: Mechanics 2.1 Motion Determine relative velocity in one and two dimensions Consider two cars, A and B, shown below. Suppose you are in car A which is moving at vA = +20 m s-1 and next to you is car B, moving at vB = +40 m s-1. As far as you are concerned your velocity vAB relative to car B is -20 m s-1 because you seem to be moving backwards relative to Bs coordinate system. We write vAB = vA - vB velocity of A relative to B

B A Determine relative velocity in one and two dimensions The equation works even in two dimensions. Suppose you are in car A which is moving at vA = +40 m s-1 and approaching you at right angles is a car B is moving at vB = -20 m s-1 as shown. Since A and B are moving perpendicular to one another, use a vector diagram to find vAB. The solution is on the next slide. x A B Topic 2: Mechanics 2.1 Motion y

B Topic 2: Mechanics 2.1 Motion Determine relative velocity in one and two dimensions Draw in the vectors and use vAB = vA - vB. v AB = vA - vB vAB2 = 402 + 202 vB -v B vA A

vAB2 = vA2 + vB2 vAB = 45 m s-1 Topic 2: Mechanics 2.1 Motion Projectile motion A projectile is an object that has been given an initial velocity by some sort of short-lived force, and then moves through the air under the influence of gravity. Baseballs, stones, or bullets are all examples of projectiles executing projectile motion. You know that all objects moving through air feel an air resistance (recall sticking your hand out of the window of a moving car). FYI We will ignore air resistance in the discussion that follows Topic 2: Mechanics

2.1 Motion ay = -g Speeding up in -y dir. ay = -g Slowing down in +y dir. Analysing projectile motion Regardless of the air resistance, the vertical and the horizontal components of velocity of an object in projectile motion are independent. Constant speed in +x dir. ax = 0 Topic 2: Mechanics 2.1 Motion Analysing projectile motion The trajectory of a projectile in the absence of air is parabolic. Know this!

Topic 2: Mechanics 2.1 Motion Analysing projectile motion with fluid resistance If there is air resistance, it is proportional to the square of the velocity. Thus, when the ball moves fast its deceleration is greater than when it moves slow. Peak to left of original one. Pre-peak distance more than postpeak. SKETCH POINTS Topic 2: Mechanics 2.1 Motion Analysing projectile motion Recall the kinematic equations: kinematic equations 1D a is constant Since we worked only in 1D at the time, we didnt have

to distinguish between x and y in these equations. Now we appropriately modify the above to meet our new requirements of simultaneous equations: s = ut + (1/2)at 2 v = u + at Displacement Velocity x = uxt + (1/2)axt 2 vx = ux + axt y = uyt + (1/2)ayt 2 vy = uy + ayt kinematic equations 2D ax and ay are constant Topic 2: Mechanics 2.1 Motion Analysing projectile motion

x 0 = uxt + (1/2)axt 2 0 vx = ux + axt y = uyt + (1/2)ayt 2 vy = uy + ayt kinematic equations 2D ax and ay are constant PRACTICE: Show that the reduced equations for projectile motion are x = uxt y = uyt - 5t 2 reduced equations of vx = ux vy = uy - 10t

projectile motion SOLUTION: ax = 0 in the absence of air resistance. ay = -10 in the absence of air resistance. Topic 2: Mechanics 2.1 Motion Analysing projectile motion x = uxt y = uyt - 5t 2 vx = ux vy = uy - 10t reduced equations of projectile motion EXAMPLE: Use the reduced equations above to prove that projectile motion is parabolic in nature. SOLUTION: Just solve for t in the first equation and substitute it into the second equation. x = uxt becomes t = x / ux so that t 2 = x2 / ux2. Then since y = uyt - 5t 2, we have

y = (uy / ux)x (5 / ux2)x2. FYI The equation of a parabola is y = Ax + Bx2. In this case, A = uy / ux and B = -5 / ux2. Topic 2: Mechanics 2.1 Motion Analysing projectile motion x = uxt y = uyt - 5t 2 vx = ux vy = uy - 10t reduced equations of projectile motion PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15. (a) What are ux and uy? SOLUTION: Make a velocity triangle. -1 s

m uy = u sin 6 5 = u = 15 uy = 56 sin 15 ux = u cos uy = 15 m s-1. ux = 56 cos 15 ux = 54 m s-1 Topic 2: Mechanics 2.1 Motion Analysing projectile motion x = uxt y = uyt - 5t 2 vx = ux vy = uy - 10t reduced equations of

projectile motion PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15. (b) What are the tailored equations of motion? (c) When will the ball reach its maximum height? SOLUTION: (b) Just substitute ux = 54 and uy = 15: x = 54t vx = 54 y = 15t - 5t2 vy = 15 - 10t tailored equations for this particular projectile (c) At the maximum height, vy = 0. Why? Thus vy = 15 - 10t becomes 0 = 15 - 10t so that Topic 2: Mechanics 2.1 Motion Analysing projectile motion

x = 54t y = 15t - 5t 2 vx = 54 vy = 15 - 10t tailored equations for this particular projectile PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15. (d) How far from the muzzle will the ball be when it reaches the height of the muzzle at the end of its trajectory? SOLUTION: From symmetry tup = tdown = 1.5 s so t = 3.0 s. Thus x = 54t x = 54(3.0) x = 160 m. Topic 2: Mechanics 2.1 Motion

Analysing projectile motion x = 54t y = 15t - 5t 2 vx = 54 vy = 15 - 10t tailored equations for this particular projectile PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15. (e) Sketch the following graphs: a vs. t, vx vs. t, vy vs. t: a SOLUTION: The only acceleration y is g in the t -10 y-direction. vx

vx = 54, a constant. Thus it does 54 t not change over time. vy = 15 - 10t Thus it is linear with vy 15 a negative gradient and it crosses t 1.5 Topic 2: Mechanics 2.1 Motion Analysing projectile motion The acceleration is ALWAYS g for projectile motionsince it is caused by Earth and its field. At the maximum height the projectile switches from upward to downward motion. v = 0 at switch. Topic 2: Mechanics 2.1 Motion Analysing projectile motion The flight time is

limited by the y motion. The maximum height is limited by the y motion. Topic 2: Mechanics 2.1 Motion Analysing projectile motion ax = 0. ay = -10 ms-2. Topic 2: Mechanics 2.1 Motion Analysing projectile motion y = uyt - 5t 2 -33 = 0t - 5t 2 -33 = -5t 2 (33/5) = t 2 Fall time limited by y-equations:

t = 2.6 s. Topic 2: Mechanics 2.1 Motion Analysing projectile motion x = uxt x = 18(2.6) Use x-equations and t = 2.6 s: x = 47 m. Topic 2: Mechanics 2.1 Motion Analysing projectile motion 18 vx = ux vx = 18. vy = uy 10t

vy = 0 10t vy = 10(2.6) = -26. tan = 26/18 = tan-1(26/18) = 55. 26 Topic 2: Mechanics 2.1 Motion Analysing projectile motion The horizontal component of velocity is vx = ux which is CONSTANT. The vertical component of velocity is vy = uy 10t , which is INCREASING (negatively). Topic 2: Mechanics 2.1 Motion Analysing projectile motion EK + EP = 0

EK = -EP EK = -mgh EKo = (1/2)mu2 EK = -(0.44)(9.8)(-32) = +138 J = EK EKo EK = +138 + (1/2)(0.44)(222) = 240 J. Topic 2: Mechanics 2.1 Motion Analysing projectile motion If 34% of the energy is consumed, 66% remains. 0.66(240) = 158 J (1/2)(0.44)v2 = 158 J v = 27 ms-1. (1/2)mvf2 - (1/2)mv2 = -EP mvf2 = mv2 + -2mg(0-H) Topic 2: Mechanics vf2 = v2 + 2gH

2.1 Motion Analysing projectile motion Use EK + EP = 0. Topic 2: Mechanics 2.1 Motion Analysing projectile motion uy = u sin uy = 28 sin 30 ux = u cos ux = 28 cos 30 ux = 24 m s-1. uy = 14 m s-1. Topic 2: Mechanics 2.1 Motion Analysing projectile motion x = uxt 16 = 24t

t = 16 / 24 = 0.67 The time to the wall is found from x y = uyt 5t 2 y = 14t 5t 2 y = 14(0.67) 5(0.67)2 = 7.1 m. Topic 2: Mechanics 2.1 Motion Analysing projectile motion 0.5s 0.0s 4m ux = x / t = (4 - 0) / (0.5 - 0.0) = 8 ms-1. Topic 2: Mechanics 2.1 Motion Analysing projectile motion 0.5s 11 m

0.0s 4m uy = y / t = (11 - 0) / (0.5 - 0.0) = 22 ms-1. Topic 2: Mechanics 2.1 Motion Analysing projectile motion 2.0 2.5 3.0 1.5 D 1.0 30 m 0.5s 11 m = tan-1(30/24) = 51 0.0s 24 m 4m

D2 = 242 + 302 so that D = 38 m ,@ = 51. Topic 2: Mechanics 2.1 Motion Analysing projectile motion New peak below and left. Pre-peak greater than postpeak.

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