Linear and Projectile Motion Chapters 2 3 Dr

Linear and Projectile Motion (Chapters 2 -3) Dr. Walker

Relative Motion • Picture yourself riding a bike alongside another person • If both of you are riding at the same speed, does your position change relative to each other? http: //keyassets. timeincuk. net/inspirewp/live/wp-content/uploads/sites/2/2017/01/St. Neots. CC 124 -630 x 420. jpg

Relative Motion • Refers to the motion of one object versus that of another • Another example – A car pulls away from your car on the interstate. You are moving, but the car in front of you is moving at a faster speed.

It’s All Relative • Although you may be at rest relative to Earth’s surface, you’re moving about 100, 000 km/h relative to the sun.

Need For Speed • Speed = distance/time • Any combination of distance and time will work

Different Types of Speed • Instantaneous Speed – The speed you are going at any given instant – When driving, you do not always go the same speed • Slowing down in a school zone for example • Average Speed – The total distance divided by the total time

Calculating Speed • Usain Bolt holds the world record for the 100 meter dash with a time of 9. 58 s. What is his speed?

Calculating Speed • Usain Bolt holds the world record for the 100 meter dash with a time of 9. 58 s. What is his speed? • 100 m / 9. 58 s = 10. 4 m/s

Calculating Speed • In training for a race, you aim to run 4. 4 m/s (about 10 mph). If you run this pace for 30 minutes, how far did you travel in meters?

Calculating Speed • In training for a race, you aim to run 4. 4 m/s (about 10 mph). If you run this pace for 30 minutes, how far did you travel in meters? • 30 minutes x 60 sec/1 min x 4. 4 m/s = 7920 m (about 5 miles)

Calculating Speed • If a cheetah can maintain a constant speed of 25 m/s, it will cover 25 meters every second. At this rate, how far will it travel in 10 seconds? In 1 minute?

Calculating Speed • If a cheetah can maintain a constant speed of 25 m/s, it will cover 25 meters every second. At this rate, how far will it travel in 10 seconds? In 1 minute? • In 10 s the cheetah will cover 250 m, and in 1 min (or 60 s) it will cover 1500 m (almost a mile!)

Velocity • Velocity is speed in a given direction – Stating a car travels at 70 mph refers only to speed • Speed is a scalar quantity – Stating a car travels east at 70 mph refers to its velocity since a direction is given • Velocity is a vector quantity • Velocity is motion in a straight line https: //boing. net/filesroot/141_velociraptor. jpg

Changing Velocity • Since velocity must have a direction, a change in speed OR direction indicates a change in velocity • These race cars have a constantly changing velocity, since they are always changing direction https: //www. gannett-cdn. com/-mm-/d 33 c 5184 df 7 eba 7194819 e 6498 d 040 da 8343774 d/c=0 -0 -39702985&r=x 404&c=534 x 401/local/-/media/2016/10/22/USATODAY/636127442487283566 talladega-start. jpg

Food For Thought • The speedometer of a car moving northward reads 60 km/h. It passes another car that travels southward at 60 km/h. Do both cars have the same speed? Do they have the same velocity?

Speed vs. Velocity The speedometer of a car moving northward reads 60 km/h. It passes another car that travels southward at 60 km/h. Do both cars have the same speed? Do they have the same velocity? Answer: Both cars have the same speed, but they have opposite velocities because they are moving in opposite directions.

Graphing Velocity • Velocity is the change in displacement divided by the change in time. – A straight-line, position-time graph indicates constant velocity. – The slope of a displacement-time graph is the velocity.

Acceleration • Acceleration is the rate at which the velocity is changing – If your velocity is constant, then your acceleration = 0!! – Acceleration is directional (like velocity), so it is a vector quantity a = Dv/Dt

Example • A Lamborghini Huracán can travel from 0 km/h to 96 km/h (roughly 0 – 60 mph) in 2. 5 s. • What is the acceleration of this vehicle?

Example • A Lamborghini Huracán can travel from 0 km/h to 96 km/h (roughly 0 – 60 mph) in 2. 5 s. • What is the top acceleration of this vehicle? • (96 km/h – 0 km/h) / 2. 5 s = 38. 4 km/h. s • Acceleration units are velocity (km/h) per time (s). Most typically, it will be m/s. s or m/s 2

Example • A car can decelerate at roughly 7 m/s 2 on a dry road. Give the time it takes a car to go from 22 m/s (roughly 55 mph) to a standstill with this rate of deceleration.

Example • A car can decelerate at roughly 7 m/s 2 on a dry road. Give the time it takes a car to go from 22 m/s (roughly 55 mph) to a standstill with this rate of deceleration. • We have to rearrange the equation to time = (change of velocity)/acceleration • (-22 m/s) / (-7 m/s 2 ) = 3. 14 s – Why are those numbers negative? • Deceleration is negative acceleration – you’re slowing down!

Example • A car is travelling at a rate of 20 m/s, then accelerates at a rate of 2 m/s 2 for 5 seconds. What is the new velocity of the car?

Example • A car is travelling at a rate of 20 m/s, then accelerates at a rate of 2 m/s 2 for 5 seconds. What is the new velocity of the car? • We can calculate the change in velocity from a = Dv/Dt. Dv = (2 m/s 2)(5 s) = 10 m/s Now we have to add this change in speed to the initial speed. 10 m/s + 20 m/s = 30 m/s This is represented by the equation Vf = Vi + at

Graphing Acceleration On a speed-versus-time graph, the speed v of a freely falling object can be plotted on the vertical axis and time t on the horizontal axis . The slope of the graph at any point is the acceleration

Free Falling Imagine there is no air resistance and that gravity is the only thing affecting a falling object. – An object moving under the influence of the gravitational force only is said to be in free fall. – The elapsed time is the time that has elapsed, or passed, since the beginning of any motion, in this case the fall.

Free Fall • During each second of free fall, the an object’s speed increases by 9. 8 meters per second. • This increase is the acceleration. • We are assuming there is no air resistance. Air resistance lowers this number, especially for objects with a large surface area relative to mass like a parachute or a feather • The letter g is used to represent this quantity. – If you see an “a” it represents the same thing

Speed of Free Fall • Instantaneous speed in free fall V = gt V = velocity, g = acceleration from gravity t = time

Example • An object is in free fall for 3. 5 s. What is it’s velocity at this time?

Example • An object is in free fall for 3. 5 s. What is it’s velocity at this time? • V = gt • V = 9. 8 m/s 2 x 3. 5 s • V = 34. 3 m/s

Free Fall – How Far? How far does an object in free fall travel? – In the first second, the object’s speed increases from 0 m/s to 10 m/s. • Average speed is 5 m/s • Travels 5 m (in 1 second) – In the second, the object’s speed increases from 10 m/s to 20 m/s • Average speed is 15 m/s, travels 15 meters – In the third second, the object’s speed increases from 20 m/s to 30 m/s. • Average speed is 25 m/s, travels 25 meters • Total distance after 3 seconds is 5 + 15 + 25 = 45 meters

Free Fall – How Far? Time (seconds) 1 2 3 4 5 Avg. velocity at that second 5 m/s 15 m/s 25 m/s 35 m/s 45 m/s Total Distance (m) 5 m 20 m 45 m 80 m 125 m There is a pattern here and it is represented by the formula d=(1/2)gt 2 D = distance, g = acceleration from gravity, t = time

Free Fall Example • Luke drops a pile of roof shingles from the top of a roof located 8. 52 meters above the ground. Determine the time required for the shingles to reach the ground.

Free Fall Example • Luke drops a pile of roof shingles from the top of a roof located 8. 52 meters above the ground. Determine the time required for the shingles to reach the ground. • Notice we’re solving for time, so we have to rearrange the equation. We’re not looking for distance very often. • D = 1/2 gt 2 8. 52 = ½(9. 8)t 2 8. 52 = 4. 9 x t 2 1. 74 = t 2 1. 32 = t

Free Fall Example • A basketball player has a vertical leap of 1. 0 m (39 inches). Calculate his “hang time”.

Free Fall Example • A basketball player has a vertical leap of 1. 0 m (39 inches). Calculate his “hang time”. • Again, we’re solving for time rather than distance • D = 1/2 gt 2 1. 0 m = ½(9. 8)t 2 0. 20 = t 2 0. 45 = t The trick here is that this is the time it takes for the player to go up 1. 0 m. It will take just as long for him to come down, so this number is multiplied by 2!! 2 x 0. 45 s = 0. 90 s

More Free Fall • Find the speed required to throw a ball straight up and have it return 5 s later. • (Hint: this is similar to the hang time example)

More Free Fall • Find the speed required to throw a ball straight up and have it return 5 s later. • The ball will travel up for ½ the time and travel down for ½ the time, so our time is 2. 5 s, not 5 s! • V = gt V = (9. 8 m/s 2)(2. 5 s) = 24. 5 m/s

More Free Fall • How far up must the ball be thrown straight up and have it return 5 s later?

More Free Fall • How far up must the ball be thrown straight up and have it return 5 s later? • D = ½ gt 2 D = ½ (9. 8 m/s 2)(2. 5 s)2 D = (4. 9 m/s 2)(6. 25 s 2) D = 30. 6 m Remember, the ball is only travelling up for ½ the time!

Graphing Free Fall • Distance vs. time is parabolic (non-linear) – The slope of this graph is velocity – Velocity is NOT constant for free fall, acceleration is!

Kinematic Equations • These equations describe and represent the motion of objects – Can be used for linear or projectile motion – Include equations that have been covered – Can measure distance or velocity, acceleration must be constant • Solve with “plug and chug” technique

Kinematic Equations 1 2 3 4 • Vf = Final velocity Vi = Initial Velocity • A = Acceleration T = time (in seconds) • Notice that equation 1 is similar to v = gt – We’re not starting from rest here • Notice equation 4 are similar to d = 1/2 gt 2 – We’re not starting at rest in equation 4 http: //study. com/cimages/multimages/16/four_kinematic_equations_fixed. png

Linear Kinematics – Example 1 • Ima Hurryin is approaching a stoplight moving with a velocity of +30. 0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8. 00 m/s 2, then determine the displacement of the car during the skidding process.

Linear Kinematics – Example 1 • Ima Hurryin is approaching a stoplight moving with a velocity of +30. 0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8. 00 m/s 2, then determine the displacement of the car during the skidding process. • What are we trying to find? • What equation should we use?

Linear Kinematics – Example 1 • Ima Hurryin is approaching a stoplight moving with a velocity of +30. 0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8. 00 m/s 2, then determine the displacement of the car during the skidding process. • What are we trying to find? Total distance • What equation should we use? vf 2 = vi 2 + 2 • a • d

Linear Kinematics – Example 1 • Ima Hurryin is approaching a stoplight moving with a velocity of +30. 0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8. 00 m/s 2, then determine the displacement of the car during the skidding process. • vf 2 = vi 2 + 2 • a • d • 0 = (+30. 0 m/s)2 + 2 • (-8. 00 m/s 2) • d = 56. 3 m

Linear Kinematics – Example 2 • Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6. 00 m/s 2 for a time of 4. 10 seconds. Determine the displacement of Ben's car during this time period. • What do you have? • What are you trying to find? • What equation do you need?

Linear Kinematics – Example 2 • Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6. 00 m/s 2 for a time of 4. 10 seconds. Determine the displacement of Ben's car during this time period. • What do you have? Acceleration, time • What are you trying to find? Total displacement • What equation do you need? d = vi • t + ½ • a • t 2

Linear Kinematics – Example 2 • Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6. 00 m/s 2 for a time of 4. 10 seconds. Determine the displacement of Ben's car during this time period. • d = vi • t + ½ • a • t 2 • Notice the first term goes away and looks like d = 1/2 gt 2 • d = 0 + ½ • 6. 00 m/s 2 • 4. 10 s = 50. 4 m

Review – Vectors vs. Scalars • Scalar quantity – Has a magnitude only – Can be manipulated like normal numbers • Vector quantity – Has magnitude AND direction – Direction has to be accounted for when manipulated – Velocity and Acceleration are vector quantities – We didn’t worry about this last chapter – we only dealt with straight line paths

Vector Diagrams • Vector quantities are often represented by scaled vector diagrams. They depict a vector by use of an arrow drawn to scale in a specific direction

Vector Diagrams • Vector diagrams should have: – a scale is clearly listed – a vector arrow (with arrowhead) is drawn in a specified direction. The vector arrow has a head and a tail. – the magnitude and direction of the vector is clearly labeled. In this case, the diagram shows the magnitude is 20 m and the direction is East.

Velocity Vectors The airplane’s velocity relative to the ground depends on the airplane’s velocity relative to the air and on the wind’s velocity.

Velocity Vectors • What if there is a crosswind?

Vector Components • A ball’s velocity can be resolved into horizontal and vertical component – In this case the ball doesn’t just go up or across, but a combination of both – Horizontal – rolling a ball on the floor – Vertical – free falling object

Velocity Vectors • The plane’s speed relative to the ground can be found by adding the two vectors. • The result of adding these two vectors, called the resultant, is the diagonal of the rectangle described by the two vectors.

Vector Diagrams • Things to notice – The lengths of the arrows are proportional to the speed – The resultant is drawn as the diagonal to the resulting rectangle – The resultant quantity comes from the pythagorean theorem (a 2 + b 2 = c 2) for right triangles

What’s Your Angle? • The direction of the resulting velocity can be determined using a trigonometric function (yes, trig!) • Since the plane velocity and the wind velocity form a right triangle when added together in head-to-tail fashion, the angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions. • In this case, the tangent is used. http: //www. physicsclassroom. com/class/vectors/Lesson-1/Relative-Velocity-and-Riverboat. Problems

On The River • A motorboat traveling 4 m/s, East encounters a current traveling 7. 0 m/s, North. – What is the resultant velocity of the motorboat? – At what angle does it travel up the river? – If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? – What distance downstream does the boat reach the opposite shore?

On The River • A motorboat traveling 4 m/s, East encounters a current traveling 7. 0 m/s, North. – What is the resultant velocity of the motorboat? – Use the Pythagorean theorem to find the length of the resultant • (4. 0)2 + (7. 0)2 = C 2 • C = 8. 06 m/s

On The River • A motorboat traveling 4 m/s, East encounters a current traveling 7. 0 m/s, North. – At what angle does it travel up the river? – q = tan(opposite/adjacent) – q = tan (7/4) – q = 60 o 7. 0 m/s q 4. 0 m/s

On The River • A motorboat traveling 4 m/s, East encounters a current traveling 7. 0 m/s, North. – If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? – Time = Distance/Speed – (80 m)(4. 0 m/s) = 20 s • Why did we did 4. 0 m/s instead of the resultant? – Width of the river – we’re only using the horizontal vector, not the resultant

On The River • A motorboat traveling 4 m/s, East encounters a current traveling 7. 0 m/s, North. – What distance downstream does the boat reach the opposite shore? – Distance = Speed x Time – Distance = 7. 0 m/s x 20 s = 140 m – We used the time from the previous step and the “vertical” vector to determine distance downstream.

Vector Jokes

Projectile Motion • Projectile – Any object that moves through the air or space, acted on only by gravity (and air resistance if present) – Examples • A bullet being fired from a gun • A ball being thrown • A car (or anything else) rolling off of a cliff

Projectile Motion • Linear motion involves motion in a straight path (linear…. line) • Projectile motion involves motion on a curved path – A curve has multiple directions • Horizontal • Vertical

Components of Projectile Motion • Horizontal – Constant velocity, no force of gravity • Vertical – Downward acceleration due to gravity • Increasing velocity • Greater distance covered each second an object falls • These components are independent of one another

Projectile Motion Demonstrated • A strobe-light photo of two balls released simultaneously–one ball drops freely while the other one is projected horizontally. Notice: 1) The horizontal motion is constant – there is no acceleration 2) Both balls cover the same vertical distance per unit time 3) This path is a parabola

A Thought Experiment • A cannonball is fired at the same time another is dropped from the same height. Which one hits the ground first?

A Thought Experiment • A cannonball is fired at the same time another is dropped from the same height. Which one hits the ground first? • Answer: Trick question! Gravity acts on both equally. The horizontal motion of the fired cannonball has no effect on vertical motion, just like the strobe light example

Projectiles At An Angle • With no gravity the projectile would follow the straightline path (dashed line). – Because of gravity it falls beneath this line the same vertical distance it would fall if it were released from rest.

Projectiles Launched At An Angle • The dashed lines show the path of the cannonball without gravity • The vertical distance it falls is the same as if it were dropped from rest

Projectiles Launched At An Angle • From a vector standpoint – Horizontal component is always the same – The vertical component constantly changes – The velocity is the resultant of the two vectors at any point in time • At the apex (top of the curve), there is no vertical component – As a result, the velocity is lowest at the top of the curve

Projectiles Launched At An Angle • The velocity of a projectile is shown at various points along its path. Notice that the vertical component changes while the horizontal component does not. Air resistance is neglected. • Note the velocity is constantly changing, because it is a vector quantity

Projectiles Launched At An Angle • Range – The distance that a projectile travels • For example, the range of ammunition – The angle at which the projectile is launched affects the distance that it travels – Notice the difference between the horizontal velocity bectors below

Projectile Launched At An Angle • Projectiles that are launched at different angles not only have different ranges, but reach different heights above the ground – This is shown by the differences in the vertical vectors

Paths of Projectiles

Paths of Projectiles • Notice in the diagram which angles have the same range – Angles that add to 90 ° have the same range – Maximum range occurs when a projectile is launched at 45 °

Speed of Projectiles • Without air resistance, a projectile will reach maximum height in the same time it takes to fall from that height to the ground. • The deceleration due to gravity going up is the same as the acceleration due to gravity coming down. • The projectile hits the ground with the same speed it had when it was projected upward from the ground.

Speed of Projectiles • Time going up = time coming down • Speed lost going up = Speed gained coming down

Projectile Problems • Remember, there is a horizontal and vertical component to a projectile • Horizontally… – How far? D = v • t – How fast? V = d/t • Vertically…. – How far? D = (1/2) • g • t 2 – How fast? D = g • t – Here g = acceleration due to gravity. We differentiate it from “a” because it’s specific to gravity

Horizontally Launched Projectiles • A pool ball leaves a 0. 60 -meter high table with an initial horizontal velocity of 2. 4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

Horizontally Launched Projectiles • A pool ball leaves a 0. 60 -meter high table with an initial horizontal velocity of 2. 4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. – What information do we have? • Horizontally, v = 2. 4 m/s, a = 0, looking for d • Vertically g = 9. 8 m/s 2, vi = 0 m/s, d = 0. 60 m, looking for t

Horizontally Launched Projectiles • A pool ball leaves a 0. 60 -meter high table with an initial horizontal velocity of 2. 4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. – What equations do we use? – For horizontal distance: x = vix • t +0. 5 • g • t 2 – For time required to drop: y = viy • t +0. 5 • g • t 2

Horizontally Launched Projectiles • A pool ball leaves a 0. 60 -meter high table with an initial horizontal velocity of 2. 4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. – What equations do we use? – For time required to drop: y = viy • t +0. 5 • g • t 2 • The vertical velocity isn’t mentioned – it’s zero so that part cancels out • 0. 60 m = 0. 5 x 9. 8 m/s 2 x t 2 • t = 0. 35 s

Horizontally Launched Projectiles • A pool ball leaves a 0. 60 -meter high table with an initial horizontal velocity of 2. 4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. – What equations do we use? – For horizontal distance: x = vix • t +0. 5 • g • t 2 • Since there no horizontal acceleration, the second term = 0. • X = (2. 4 m/s)(0. 35 s) =0. 84 m

Another Example • A soccer ball is kicked horizontally off a 22. 0 -meter high hill and lands a distance of 35. 0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

Another Example • A soccer ball is kicked horizontally off a 22. 0 -meter high hill and lands a distance of 35. 0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. – – Vertical first – need to determine time y = viy • t +0. 5 • g • t 2 22. 0 m = 0. 5 (9. 8 m/s 2) (t 2) t = 2. 1 s

Another Example • A soccer ball is kicked horizontally off a 22. 0 -meter high hill and lands a distance of 35. 0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball. – – Determine velocity with horizontal x = vix • t + 0. 5 • g • t 2 35. 0 m = v (2. 1 s) v = 16. 7 m/s