PHY 151: Lecture 2 Study of Motion 2.1 Average Velocity

PHY 151: Lecture 2 Study of Motion 2.1 Average Velocity
2.2 Instantaneous Velocity 2.3 Particle under Constant Velocity 2.4 Acceleration 2.5 Motion Diagrams 2.6 Particle Under Constant Acceleration 2.7 Freely Falling Objects 2.8 Context Connection: Acceleration Required by Consumers

Study of Motion - 1 Everything in the universe is moving
To understand the universe, we must understand motion Historically, motion was the first classical physics topic that was understood

Study of Motion - 2 Mechanics Branch of physics that deals with motion
Kinematics Description of motion Dynamics What causes motion or changes motion

PHY 151: Lecture 2 Motion in One Dimension
2.1 Average Velocity

Average Velocity - 1 Kinematics describes motion while ignoring the agents that caused the motion For now, we will consider motion in one dimension Along a straight line We will use the particle model A particle is a point-like object, has mass but infinitesimal size

Average Velocity - 2 Average speed is defined as
Speed is not a vector, so has no direction Average velocity is a vector quantity

Average Velocity - 3 The motion of a particle can be specified if its position is known at all times Consider the car moving back and forth along the x axis The figure is a pictorial representation of the motion Imagine we take data on the position of the car every 10 s

Average Velocity – 4 The graphical representation of the motion is a position-time graph The smooth curve is a guess as to what happened between the data points

Average Velocity – 5 The tabular representation of the motion is shown here The data are entries for position at each time

Average Velocity - 6 Average velocity: defined as ratio of displacement x to time interval t Subscript x indicates motion along x axis Dimensions: m/s (SI) x can be positive or negative

Average Velocity – 7 Average velocity can be interpreted geometrically
A straight line can be drawn between any two points on curve Line forms the hypotenuse of right triangle of height x and base t Slope of hypotenuse is ratio x/t Average velocity during time interval t is slope of the line joining initial and final points on the position-time graph

Average Velocity - 8 The displacement of a particle during the time interval ti to tf is equal to the area under the curve between the initial and final points on a velocity-time curve

Example 2.1 Find the displacement, average velocity, and average speed of the car between positions A and F. Find the displacement of the car: Find the car’s average velocity:

Example 2.1 Find the car’s average speed:
We cannot unambiguously find the average speed of the car from the data in Table 2.1, because we do not have information about the positions of the car between the data points Assume the distance from A to B is 22 m. The distance from B to F is 105 m for a total o 127 m distance.

Example 2.2 A jogger runs in a straight line, with an average velocity of magnitude 5.00 m/s for 4.00 min and then with an average velocity of magnitude 4.00 m/s for 3.00 min.

Example 2.2 (A) What is the magnitude of the final displacement from her initial position? Find the displacement for each portion:

Example 2.2 (B) What is the magnitude of her average velocity during this entire time interval of 7.00 min? Find the average velocity for the entire time interval:

2.2 Instantaneous Velocity

Instantaneous Velocity - 1
Instantaneous velocity is the limit of the average velocity as the time interval becomes infinitesimally short or as the time interval approaches zero The instantaneous velocity indicates what is happening at every point of time

The instantaneous velocity is the slope of the line tangent to the x vs. t curve the green line in the figure The blue lines show that as t gets smaller, they approach the green line

The general equation for instantaneous velocity is dx/dt is the derivative of x with respect to t The instantaneous velocity can be positive, negative, or zero

Instantaneous speed: magnitude of the instantaneous velocity vector Speed can never be negative

Example 2.3 The position of a particle moving along the x axis varies in time according to the expression x = 3t2, where x is in meters and t is in seconds. Find the velocity in terms of t at any time. The initial coordinate at time t is xi = 3t2, the coordinate at a later time t + t is Find the displacement in the time interval t:

Example 2.3 Find the average velocity in this time interval:
To find the instantaneous velocity, take the limit as t approaches zero: At t = 3 s, vx = 18 m/s

Example 2.4 A particle moves along the x axis. Its position varies with time according to the expression x = 4t + 2t2, where x is in meters and t is in seconds. Notice that the particle moves in the negative x direction for the first second of motion, is momentarily at rest at the moment t = 1 s, and moves in the positive x direction at times t > 1 s.

Example 2.4 (A) Determine the displacement of the particle in the time intervals t = 0 to t = 1 s and t = 1 s to t = 3 s. In the first time interval, set ti = tA = 0 and tf = tB = 1 s, find the displacement: For the second time interval (t = 1 s to t = 3 s), set ti = tB = 1 s and tf = tD = 3 s: These displacements can also be read directly from the position–time graph.

Example 2.4

2.3 Particle Under Constant Velocity

Particle Under Constant Velocity - 1
If the velocity of a particle is constant: Its instantaneous velocity at any instant is the same as the average velocity over a given time period: vx = vx, avg = Dx/Dt xf = xi + vxt These equations can be applied to particles or objects that can be modeled as a particle moving under constant velocity

Example – 2.5 A kinesiologist is studying the biomechanics of the human body. (Kinesiology is the study of the movement of the human body. Notice the connection to the word kinematics.) She determines the velocity of an experimental subject while he runs along a straight line at a constant rate. The kinesiologist starts the stopwatch at the moment the runner passes a given point and stops it after the runner has passed another point 20 m away. The time interval indicated on the stopwatch is 4.0 s.

Example – 2.5 (A) What is the runner’s velocity?
Find the constant velocity of the runner: (B) If the runner continues his motion after the stopwatch is stopped, what is his position after 10 s has passed? Find the position of the particle (runner) at time t = 10 s:

2.4 Acceleration

Acceleration - 1 Acceleration is the rate of change of the velocity
It is a measure of how rapidly the velocity is changing Dimensions are L/T2 SI units are m/s2

Acceleration – 2 Instantaneous acceleration is the limit of the average acceleration as Dt approaches zero: Because vx = dx/dt, this can also be written as

Acceleration – 3 The slope of the velocity vs. time graph is the acceleration Positive values correspond to where velocity in the positive x direction is increasing The acceleration is negative when the velocity is in the positive x direction and is decreasing

Acceleration – 4 Negative acceleration does not necessarily mean that an object is slowing down If the velocity is negative and the acceleration is negative, the object is speeding up Deceleration is commonly used to indicate an object is slowing down, but will not be used in this text When an object’s velocity and acceleration are in the same direction, the object is speeding up in that direction When an object’s velocity and acceleration are in the opposite direction, the object is slowing down

Example 2.6 The velocity of a particle moving along the x axis varies according to the expression vx = 40  5t2 where vx is in meters per second and t is in seconds.

Example 2.6 (A) Find the average acceleration in the time interval t = 0 to t = 2.0 s. Find the velocities at ti = tA = 0 and tf = tB = 2.0 s: Find the average acceleration in the specified time interval:

Example 2.6 (B) Determine the acceleration at t = 2.0 s.
Knowing that the initial velocity at any time t is vxi = 40  5t2, find the velocity at any later time t + t: Find the change in velocity over the time interval t:

Example 2.6 Divide by t and take the limit: Substitute t = 2.0 s:

2.5 Motion Diagrams Skipped

2.6 Particle Under Constant Acceleration Skipped Some

Particle Under Constant Acceleration - 1
If a particle has constant acceleration, we can use the following equations:

Example 2.7 A jet lands on an aircraft carrier at a speed of 140 mi/h ( 63 m/s). (A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the jet and brings it to a stop? Find the acceleration of the jet, modeled as a particle:

Example 2.7 (B) If the jet touches down at position xi = 0, what is its final position?

Example 2.8 A car traveling at a constant speed of 45.0 m/s passes a trooper on a motorcycle hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch the car, accelerating at a constant rate of 3.00 m/s2. How long does it take her to overtake the car?

Example 2.8 Find the position at any time t:
The trooper starts from rest at tB = 0 and accelerates at ax = 3.00 m/s2 away from the origin. Find her position at any time t:

Example 2.8 Set the positions of the car and trooper equal to represent the trooper overtaking the car at position C: Rearrange the quadratic equation:

Example 2.8 Solve the equation for time:
Evaluate the solution (choose the positive root):

1-Dimension Horizontal Motion Example 1
Basketball player starts from rest and accelerates uniformly to speed 6.0 m/s in 1.5 s What distance does the player run? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 0 xf = ? vi = 0 vf = 6.0 a = ? t = 1.5 6=0+a(1.5) Equation has 1 unknown, a (1st) xf=0+0(1.5)+½a(1.5)2 Equation has 2 unknowns, xf, a (2nd) 62=02+2a(xf–0) Equation has two unknowns, xf, a 6=0+a(1.5); a = 6/1.5 = 4 m/s2 xf=0+0(1.5)+½(4)(1.5)2 = 4.5 m/s

1-Dimension Horizontal Motion Example 2a
Car accelerates from rest at rate of 2.0 m/s2 for 5.0 s (a) What is the speed of the car at the end of that time? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 0 xf = ? vi = 0 vf = ? a = 2.0 t = 5.0 vf=0+2(5) Equation has 1 unknown, vf xf=0+0(5)+½(2)(5)2 Equation has 1 unknown, xf vf2=02+2(2)(xf–0) Equation has two unknowns, xf, vf vf=0+2(5)=10 m/s

1-Dimension Horizontal Motion Example 2b
Car accelerates from rest at rate of 2.0 m/s2 for 5.0 s (b) How far does the car travel in this time? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 0 xf = ? vi = 0 vf = 10 a = 2.0 t = 5.0 10=0+2(5) Can’t use Equation xf=0+0(5)+½(2)(5)2 Equation has 1 unknown, xf 102=02+2(2)(xf–0) Equation has 1 unknown, xf xf=0+0(5)+½(2)(5)2 = 25 m

1-Dimension Horizontal Motion Example 3a
A car traveling at 15 m/s stops in 35 m (a) What is the acceleration? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 0 xf = 35 vi = 15 vf = 0 a = ? t = ? 0=15+at Equation has 2 unknowns, a, t 35=0+15t+½(a)(t)2 Equation has 2 unknowns, a, t 02=152+2a(35–0) Equation has 1 unknown, a 02=152+2a(35–0) a = -225/70 = -3.2 m/s2

1-Dimension Horizontal Motion Example 3b
A car traveling at 15 m/s stops in 35 m (b) What is time during this deceleration until car stops? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 0 xf = 35 vi = 15 vf = 0 a = -3.2 t = ? 0=15-3.2t Equation has 1 unknown, t 35=0+15t+½(-3.2)(t)2 Equation has 1 unknown, t 02=152+2(-3.2)(35–0) Can’t use equation 0=15-3.2t t = -15/-3.2 = 4.67 s

A plane accelerates at 8 m/s2 on a runway that is 500 m long The take off speed of the plane is 80 m/sec Can the plane takeoff? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 0 xf = 500 vi = 0 vf = ? a = 8 t = ? vf=0+8t Equation has 2 unknowns, vf, t 500=0+0t+½(8)(t)2 Equation has 1 unknown, t vf2=02+2(8)(500–0) Equation has 1 unknown, vf vf2=02+2(8)(500–0) = 500(16) vf=sqrt(500(16) = 89.4 m/s The plane can takeoff

A plane accelerates at 8 m/s2 The take off speed of the plane is 80 m/sec What is minimum length of run way for plane to reach take off speed? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 0 xf = ? vi = 0 vf = 80 a = 8 t = ? 80=0+8t Equation has 1 unknown, t xf=0+0t+½(8)(t)2 Equation has 2 unknowns, xf, t 802=02+2(8)(xf–0) Equation has 1 unknown, xf 802=02+2(8)(xf–0) xf=6400/16 = 400 m

1-Dimension Horizontal Motion Example 6 Bull
Bull runs 8 m/sec A boy at rest has a head start of 12 m When he sees the bull he accelerates at 2 m/s2 Does the bull catch the boy? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 0 xf = ? vi = 8 vf = 8 a = 0 t = ? 8=8+0t Can’t use this equation xf=0+8t+½(0)(t)2 Equation has 2 unknowns xf, t 82=82+2(0)(xf–0) Can’t use this equation

1-Dimension Horizontal Motion Example 6 Boy
Bull runs 8 m/sec A boy at rest has a head start of 12 m When he sees the bull he accelerates at 2 m/s2 Does the bull catch the boy? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 12 xf = ? vi = 0 vf = ? a = 2 t = ? vf=0+2t Equation has 2 unknowns, vf, t xf=12+0t+½(2)(t)2 Equation has 2 unknowns, xf, t vf2=02+2(2)(xf–12) Equation has 2 unknowns, xf, vf

1-Dimension Horizontal Motion Example 6 Bull / Boy
Bull runs 8 m/sec A boy at rest has a head start of 12m When he sees the bull he accelerates at 2 m/s2 Does the bull catch the boy? xf=0+8t+½(0)(t)2 (from Bull equations) xf=12+0t+½(2)(t)2 (from Boy equations) Bull catches boy when their xf are equal 8t=12+t2 t2-8t+12=0 Factor (t-2)(t-6) = 0 Bull catches boy at 2 s and 6 s 2 s is bull catching up to the boy 6 s is the accelerating boy catching up to the bull

1-Dimension Horizontal Motion Example 7 Car
Car travels at 45.0 m/s passes a trooper One second later the trooper sets out to catch the car accelerating at 3.00 m/s2 How long does it take the trooper to overtake the car? Let t = 0 be 1 second after the car passes the trooper The cars position is then at xi = 45 m vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 45 xf = ? vi = 45 vf = 45 a = 0 t = ? 45=45+0t Can’t use this equation xf=45+45t+½(0)(t)2 Equation has 2 unknowns, xf, t 452=452+2(0)(xf–0) Can’t use this equation

1-Dimension Horizontal Motion Example 7 Trooper
Car travels at 45.0 m/s passes a trooper One second later the trooper sets out to catch the car accelerating at 3.00 m/s2 How long does it take the trooper to overtake the car? vf = vi + at xf = xi + vit + ½at2 vf2 = vi2 + 2a(xf – xi) xi = 0 xf = ? vi = 0 vf = ? a = 3 t = ? vf=0+3t Equation has 2 unknowns, vf, t xf=0+0t+½(3)(t)2 Equation has 2 unknowns, xf, t vf2=02+2(3)(xf–0) Equation has 2 unknowns, xf, vf

1-Dimension Horizontal Motion Example 7 Car / Trooper
Car travels at 45.0 m/s passes a trooper One second later the trooper sets out to catch the car accelerating at 3.00 m/s2 How long does it take the trooper to overtake the car? xf=45+45t+½(0)(t)2 (from Car equations) xf=0+0t+½(3)(t)2 (from Boy equations) Trooper catches car when their xf are equal 45+45t = 1.5t2 t2-30t-30=0 Solve using quadratic equation t = s rounded to 31.0 s

2.7 Freely Falling Objects

Freely Falling Objects - 1
Galileo performed many systematic experiments on objects moving on inclined planes With careful measurements, he showed displacement of object starting from rest is proportional to the square of the time interval

Freely Falling Objects - 2
A freely falling object is any object moving freely under the influence of gravity alone (neglecting air resistance) All objects fall with the same acceleration near Earth’s surface, 9.80 m/s2 downward A freely falling object can be: Dropped (released from rest) Thrown downward Thrown upward An object in free fall can be modeled as a particle under constant acceleration

Example 2.9 A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on its way down.

Example 2.9 (A) Using t𝖠 = 0 as the time the stone leaves the thrower’s hand at position 𝖠, determine the time at which the stone reaches its maximum height. Calculate the time at which the stone reaches its maximum height: Substitute numerical values:

Example 2.9 (B) Find the maximum height of the stone.
Set yA = 0 and substitute the time from part (A) to find the maximum height:

Example 2.9 (C) Determine the velocity of the stone when it returns to the height from which it was thrown. Choose the initial point where the stone is launched and the final point when it passes this position coming down:

Example 2.9 (D) Find the velocity and position of the stone at t = 5.00 s. Calculate the velocity: Find the position at t = 5.00 s:

1-Dimension Free Fall - Example 1
A bomb is dropped from 6000m With what speed does it hit the ground? vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = 6000 yf = 0 vi = 0 vf = ? a = t = ? vf = 0 – 9.8t Equation has 2 unknowns vf, t 0 = t + ½ (-9.8)t2 Equation has 1 unknown, t vf2 = (-9.8) (0 – 6000) Equation has 1 unknown, vf vf2 = (-9.8)(-6000) vf = sqrt(117600) = -343 m/s

1-Dimension Free Fall – Example 2a
A student drops a ball from the top of a tall building It takes 2.8 s for the ball to reach the ground (a) What is the height of the building? vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = ? yf = 0 vi = 0 vf = ? a = -9.8 t = 2.8 vf = 0 – 9.8(2.8) Equation has 1 unknown, vf 0 = yi + 0(2.8) + ½(-9.8)(2.8)2 Equation has 1 unknown, yi vf2 = (-9.8) (0 – yi) Equation has 2 unknowns, yi, vf 0 = yi + (1/2)(-9.8)(2.8)2 yi = 38.4 m

1-Dimension Free Fall – Example 2b
A student drops a ball from the top of a tall building It takes 2.8 s for the ball to reach the ground (b) What was the ball’s velocity just before hitting the ground? vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = yf = 0 vi = 0 vf = ? a = t = 2.8 vf = 0 – 9.8(2.8) Equation has 1 unknown, vf 0 = (2.8) + ½(-9.8)(2.8)2 Can’t use equation vf2 = (-9.8) (0 – 38.4) Equation has 1 unknown, vf vf = 0 – 9.8(2.8) = m/s (- means downward)

1-Dimension Free Fall – Example 3
Boy throws stone straight upward with an initial velocity of 15 m/s What maximum height will stone reach before falling back down? At the maximum height vy = 0 vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = 0 yf = ? vi = 15 vf = 0 a = t = ? 0 = 15 – 9.8t Equation has 1 unknown, t yf = t + ½(-9.8)t2 Equation has 2 unknown, yf, t 02 = (-9.8) (yf – 0) Equation has 1 unknown, yf 02 = (-9.8) (yf – 0) yf = -152/2/-9.8 = m

1-Dimension Free Fall – Example 4
Ball thrown upwards at 40m/s. Calculate time to reach 35m vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = 0 yf = 35 vi = 40 vf = ? a = t = ? vf = 40 – 9.8t Equation 2 unknown vf,t (2nd) 35 = t + ½(-9.8)t2 Equation has 1 unknown, t vf2 = (-9.8) (35 – 0) Equation 1 unknown, vf (1st) vf2 = (-9.8) (35 – 0) = 914 vf = or -30.2 30.2 = 40 – 9.8t t = (30.2 – 40) / (-9.8) = 1.0 s (time on the way up)

1-Dimension Free Fall Example 5 Ball 1
A ball is dropped from rest Four seconds later, a second ball is thrown down at 50m/s Calculate when and where the two balls meet Let t = 0 at 4 seconds. Ball height y=½(-g)t2=½(-9.8)(16) = m. Speed v = (-g)t = -9.8(4) = m/s vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = yf = ? vi = vf = ? a = t = ? vf = – 9.8t Equation has 2 unknowns, vf, t yf = – 39.2t + ½(-9.8)t2 Equation 2 unknowns yf, t vf2 = (-29.2)2+2(-9.8) (yf – (-78.4)) Equation 2 unknowns, yf, vf

1-Dimension Free Fall Example 5 Ball 2
A ball is dropped from rest Four seconds later, a second ball is thrown down at 50m/s Calculate when and where the two balls meet vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = 0 yf = ? vi = -50 vf = ? a = t = ? vf = -50–9.8t Equation has 2 unknown, vf, t yf = 0-50t+½(-9.8)t2 Equation has 2 unknowns, yf, t vf2 = 502+2(-9.8) (yf – 0) Equation has 2 unknowns, yf, vf

1-Dimension Free Fall Example 5 Balls 1 and 2
A ball is dropped from rest Four seconds later, a second ball is thrown down at 50m/s Calculate when and where two balls meet yf = – 39.2t + ½(-9.8)t2 (from ball 1) yf = 0-50t+½(-9.8)t2 (from ball 2) Balls meet when positions are the same 2 t+ ½ (-9.8)t2 = -50t + ½(-9.8)t2 0 = 10.8t -78.4 t = / = 7.3 s

1-Dimension Free Fall – Example 6a
From tower 100 m high, ball is thrown up with speed of 40 m/s (a) How high does it rise? vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = 100 yf = ? vi = 40 vf = 0 a = t = ? 0 = 40 – 9.8t Equation has 1 unknown, t yf = t + ½(-9.8)t2 Equation has 2 unknowns, yf, t 02 = 402+2(-9.8)(yf–100) Equation has 1 unknown, yf 02 = 402+2(-9.8)(yf–100) yf = (-1600/2/-9.8) = m

1-Dimension Free Fall – Example 6b
From tower 100 m high, ball is thrown up with speed of 40 m/s (b) When does it hit ground? vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = 100 yf = 0 vi = 40 vf = ? a = t = ? vf = 40 – 9.8t Equation has 2 unknowns, vf, t 0 = t + ½(-9.8)t2 Equation has 1 unknown, t vf2 = 402+2(-9.8)(0–100) Equation has 1 unknown, vf 0 = t + ½(-9.8)t2 -4.9t2 + 40t = 0 (Use quadratic equation) t = 10.2 s

1-Dimension Free Fall – Example 6c
From tower 100 m high, ball is thrown up with speed of 40 m/s (c) How fast is it moving at the ground ? vf = vi + at yf = yi + vit + ½at2 vf2 = vi2 + 2a(yf – yi) yi = 100 yf = 0 vi = 40 vf = ? a = t = 10.2 vf = 40 – 9.8(10.2) Equation has 1 unknown, vf 0 = (10.2)+½(-9.8)(10.2)2 Can’t use equation vf2 = 402+2(-9.8)(0–100) Equation has 1 unknown, vf vf2 = 402+2(-9.8)(0–100) = = 3560 vf = m/s

2.8 Context Connection: Acceleration Required by Consumers Skipped

PHY 151: Lecture 2 Study of Motion 2.1 Average Velocity

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