Kinematics IN 1D, 2D AND 3D PDF

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KINEMATICS: ONE DIMENSIONAL MOTION INTENDED LEARNING OUTCOMES (ILO) • Describe the motion of a body in terms of position, velocity, and acceleration. • Relate displacement and time to velocity and acceleration. • Apply calculus in deriving motion equations. • Interpret motion graphs. Kinematics - is the science of describing the motion of objects using words, diagrams, numbers, graphs, and equations. One dimensional motion – is a motion along a straight line. Position: the location of the particle with respect to a chosen reference point that we can consider to be the origin of a coordinate system. Distance: is a scalar quantity and it is the length of a path followed by a particle. Displacement: - is a vector quantity and it is the change between the starting point and final point.

∆𝒙 𝒅𝒙 = = 𝒗(𝒕) ∆𝒕→𝟎 ∆𝒕 𝒅𝒕

𝒗𝒙 = 𝐥𝐢𝐦

Average acceleration: the change in velocity divided by the time interval during which that change occurs.

󰇍 𝒂𝒗𝒆 = 𝒂

∆𝒗 𝒗𝒇 − 𝒗𝒊 = ∆𝒕 𝒕𝒇 − 𝒕𝒊

Instantaneous acceleration: the limit of the average acceleration as approaches zero.

∆𝒗 𝒅𝒗 = = 𝒂(𝒕) ∆𝒕→𝟎 ∆𝒕 𝒅𝒕

𝒂𝒙 = 𝐥𝐢𝐦

𝒅𝒙 𝐝𝐯 𝒅 𝒅𝒕 𝒅𝟐 𝒙 = 𝒂= = 𝐝𝐭 𝒅𝒕 𝒅𝒕

1. A particle moves along the x-axis. Its position varies with time according to the expression (𝑡) = −4𝑡 + 2𝑡 2 , where x is in meters and t is in seconds. 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.

∆𝒙 = 𝒙𝒇 − 𝒙𝒊 Speed is the rate of change of a distance. Velocity is the rate of change of a displacement. Average speed: a scalar quantity, defined as the total distance traveled divided by the total time interval required to travel that distance.

𝒗𝒂𝒗𝒆 =

𝒅 𝒅 = ∆𝒕 𝒕

Average velocity: the particle’s displacement divided by the time interval during which that displacement occurs.

󰇍 𝒂𝒗𝒆 = 𝒗

b.) Calculate the average velocity during these two time intervals.

∆𝒙 𝒙𝒇 − 𝒙𝒊 = ∆𝒕 𝒕𝒇 − 𝒕𝒊

c.) Find the instantaneous velocity of the particle at t = 2.5 s.

Instantaneous velocity: the rate of motion of a particle or object at a given time.

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2. The velocity of a particle moving along the x-axis varies according to the expression 𝑣𝑥 = 40 − 5𝑡2 where t is in seconds. a.) Find the average acceleration in the time interval t=0s to t=2s .

KINEMATIC EQUATIONS The kinematic equations are a set of equations that can be utilized to predict unknown information about an object's motion if other information is known. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known.

b.) Determine the acceleration at t=2s. Kinematics Equation 1:

c.) Determine the average velocity in the time interval t=0s to t=2s .

Kinematics Equation 2:

Kinematics Equation 3:

3. Find the displacement, average velocity, and average speed of the car between position A and F as described by the table.

where: ∆𝑥 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡. 𝑡 = 𝑡𝑖𝑚𝑒 𝑤ℎ𝑖𝑐ℎ 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑚𝑜𝑣𝑒𝑑. 𝑎 = 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡. 𝑣𝑓 = 𝑓𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡. 𝑣𝑖 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡.

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1. A bus is moving at 25 m/s when the driver steps on the brakes and brings the bus to a stop in 3.0 s. What is the acceleration of the bus while braking?

5. 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/s^2. How long does it take her to overtake the car?

2. A golf ball rolls up a hill toward a miniature-golf hole. Assign the direction toward the hole as being positive. If the ball starts with a speed of 2.0 m/s and slows at a constant rate of 0.50 m/s2, what is its velocity after 2.0 s?

FREELY FALLING BODIES 3. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

Free fall is the motion of an object under the influence of only gravity. The velocity change is the same in each time interval, so the acceleration is constant. Aristotle thought that heavy bodies fall faster than light ones, but Galileo showed that all bodies fall at the same rate. If there is no air resistance, the downward acceleration of any freely falling object is g = 9.8 m/s2 = 32 ft/s2. An object is in free fall even when it is moving upward. Refer to the figure for up-and-down motion.

4. Lawrence is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Lawrence applies the brakes and skids to a stop. If Lawrence’s acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process and the time the car will stop.

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KINEMATIC EQUATIONS FOR FREELY FALLING BODIES 𝒗𝒇 = 𝒗𝒊 + 𝒈𝒕 (𝟏) ∆𝒚 = 𝒗𝒊 𝒕 +

𝒈𝒕𝟐 𝟐

(𝟐)

4. A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down, as shown in the figure.

𝒗𝒇 𝟐 = 𝒗𝒊 𝟐 + 𝟐𝒈∆𝒚 (𝟑) 1. 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.

2. Rex throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

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.

b. ) Find the maximum height of the stone.

3. The boy drops the ball from a roof of the house which takes 3 seconds to hit the ground. Calculate the velocity before the ball crashes to the ground. How high is the roof?

c.) Determine the velocity of the stone when it returns to the height from which it was thrown.

d.) Find the velocity and position of the stone at t = 5.00 s.

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KINEMATICS: TWO AND THREE DIMENSIONAL MOTION INTENDED LEARNING OUTCOMES (ILO)  Find the position, velocity and acceleration using unit vectors  Describe and compare the motion of a body thrown horizontally to that of in free fall  Derive working equations for horizontallylaunched projectile  Study the elements of projectile motion  Determine the relationship of the angle of projection on the range at a given initial velocity. Motion in Two Dimensions Kinematic variables in one dimension  Position: x(t) m  Velocity: v(t) m/s  Acceleration: a(t) m/s2 Kinematic variables in three dimensions  Position: 𝑟(𝑡) = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 , 𝑚  Velocity: 𝑣(𝑡) = 𝑣𝑥 𝑖 + 𝑣𝑦 𝑗 + 𝑣𝑧 𝑘 , 𝑚/𝑠  Acceleration: 𝑎(𝑡) = 𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 + 𝑎𝑧 𝑘 , 𝑚/𝑠2

Instantaneous Velocity: the derivative of the position vector with respect to time and direction at any point in a particles path is along a line tangent to the path at that point.

󰇍 𝒅𝒓 ∆𝒓 = ∆𝒕→𝟎 ∆𝒕 𝒅𝒕

󰇍𝒗 = 𝐥𝐢𝐦 󰇍𝒗 =

𝒅𝒚 𝒅𝒙 𝒊 + 𝒋 𝒅𝒕 𝒅𝒕

󰇍𝒗 = 󰇍𝒗𝒙 𝒊 + 󰇍𝒗 𝒚𝒋

Average acceleration: change in its instantaneous velocity vector divided by the time interval ∆𝑡 during which changes occurs

󰇍𝒂𝒂𝒗𝒈 =

Position vectors : the position of an object is described by its position vector - always points to particle from origin.

∆𝒗𝒚 ∆𝒗󰇍 ∆𝒗𝒙 𝒋 = 𝒊 + ∆𝒕 ∆𝒕 ∆𝒕

󰇍𝒂𝒂𝒗𝒈 = 𝒂󰇍 𝒂𝒗𝒈𝒙 𝒊 + 󰇍𝒂𝒂𝒗𝒈𝒚𝒋

Instantaneous acceleration: The derivative of the velocity vector with respect to time.

∆𝒗 𝒅𝒗 = ∆𝒕→𝟎 ∆𝒕 𝒅𝒕

󰇍𝒂 = 𝐥𝐢𝐦 𝒂 󰇍 𝒂𝒗𝒈 = 𝐥𝐢𝐦 ∆𝒕→𝟎

Displacement vector: difference between its final position vector and its initial position vector but making use of the full vector notation rather than positive and negative signs to indicate the direction of motion.

󰇍 = 󰇍󰇍󰇍 𝒓󰇍𝟐 − 𝒓󰇍󰇍󰇍𝟏 ∆𝒓 󰇍 = (𝒙𝟐𝒊 + 𝒚𝟐 𝒋) − (𝒙𝟏𝒊 + 𝒚𝟏 𝒋) ∆𝒓 ∆𝒓󰇍 = ∆𝒙𝒊 − ∆𝒚𝒋 Average velocity: the displacement vector of the particle divided by the time interval

󰇍𝒗𝒂𝒗𝒈 =

∆𝒚 ∆𝒓󰇍 ∆𝒙 = 𝒊 + 𝒋 ∆𝒕 ∆𝒕 ∆𝒕

𝒗 󰇍 𝒂𝒗𝒈 = 𝒗󰇍 𝒂𝒗𝒈𝒙 𝒊 + 𝒗 󰇍 𝒂𝒗𝒈𝒚𝒋

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󰇍𝒂 =

𝒅𝒗𝒚 𝒅𝒗𝒙 𝒋 𝒊 + 𝒅𝒕 𝒅𝒕

󰇍𝒂 = 𝒂𝒙 𝒊 + 𝒂𝒚𝒋

 The direction of the velocity can change, even though the magnitude is constant  Both the magnitude and the direction can change Motion in two dimensions Motions in each dimension are independent components

𝑣 = 𝑣𝑖 + 𝑎𝑡 1 𝑟 − 𝑟󰇍𝑖 = 󰇍𝑣󰇍 𝑖𝑡 + 𝑎𝑡2 2

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Constant acceleration equations

𝒗𝒙 = 𝒗𝒊𝒙 + 𝒂𝒙 𝒕 1 ∆𝒙 = 𝒗𝒙𝒊 𝒕 + 𝑎 𝑡 2 2 𝑥 𝒗𝒇𝒙 𝟐 = 𝒗𝒊𝒙 𝟐 + 𝟐𝑎𝑥 ∆𝑥

𝒗𝒚 = 𝒗𝒊𝒚 + 𝒂𝒚𝒕

2 1 ∆𝒚 = 𝒗𝒊𝒚 𝒕 + 2 𝑎𝑦 𝑡 𝒗𝒇𝒚𝟐 = 𝒗𝒊𝒚 𝟐 + 𝟐𝑎 𝑦 ∆𝑦

Constant acceleration equations hold in each dimension  𝑡 = 0 beginning of the process;  𝑎 = 𝑎𝑥 𝑖 + 𝑎𝑦 𝑗 where 𝑎𝑥 and 𝑎𝑦 are constant;  Initial velocity 󰇍󰇍𝑣󰇍𝑖 = 𝑣𝑖𝑥 𝑖 + 𝑣𝑖𝑦 𝑗  initial displacement 󰇍𝑟𝑖 = 𝑟𝑖 𝑖 + 𝑟𝑖 𝑗 To determine the magnitude of a vector and its direction: 󰇍 | = √𝑨𝒙𝟐 + 𝑨𝒚 𝟐 𝑨 = |𝑨 𝜽 = 𝒕𝒂𝒏−𝟏 |

𝑨𝒚 | 𝑨𝒙

2. A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by the following expressions: 𝑥 = (18.0 𝑚/𝑠)𝑡 𝑦 = (4.00 𝑚/𝑠)𝑡 − (4.90 𝑚/𝑠2 )𝑡 2 a.) Write a vector expression for the ball’s position as a function of time, using the unit vectors î and ĵ.

b.) The velocity vector as a function of time

c.) The acceleration vector as a function of time. Next use unit–vector notation to write expressions.

d.) The position, the velocity, and the acceleration of the golf ball, all at t = 3.00 s.

1. A particle starts from the origin at t = 0 with an initial velocity having an x component of 20 m/s and a y component of -15 m/s. The particle moves in the xy plane with an x component of acceleration only, given by ax=4.0 m/s2. a.) Calculate the velocity and speed of the particle at t = 5.0s. Evaluate the speed of the particle as the 󰇍󰇍󰇍𝒇. magnitude of 𝒗

b.) Determine the x and y coordinates of the particle at any time t and its position vector at this time.

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3. A web page designer creates an animation in which a dot on a computer screen has a position of 𝑟 = [4.0𝑐𝑚 + (2.5𝑐𝑚/𝑠 2 )𝑡 2 ]𝑖 + (5.0𝑐𝑚/𝑠)𝑡 𝑗 a.) Find the magnitude and direction of the dot’s average velocity between t = 0 and t =2s.

b.) Find the magnitude and direction of the dot’s instantaneous velocity at t = 0, t = 1s, and t = 2.0s.

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PROJECTILE MOTION  curved motion with constant acceleration. It is two dimensional motion of a particle thrown obliquely into the air.  is an object thrown into the air upon which the only force acting is gravity.  is any object which once projected continues in motion by its own inertia and is influenced only by the downward force of gravity  The path of the projectile is always a parabola.  The path followed by the projectile is termed as the "trajectory of the projectile". Example of Projectile Motion 1. An object thrown from a hill to the downward direction. 2. An object thrown from the ground towards the sky or in the upward direction. 3. An object thrown towards the sky but by making some angle with horizontal surface . Horizontal and Vertical Components of a Projectile’s Motion  projectile travel with a parabolic trajectory due to the influence of gravity.  there are no horizontal forces acting upon a projectile and thus no horizontal acceleration.  the horizontal velocity of a projectile is constant (considering air resistance is negligible)  there is a vertical acceleration caused by gravity; its value is 9.8 m/s2, downward.  the vertical velocity of a projectile changes by 9.8 m/s each second.  the horizontal and vertical components of a projectile motion are independent of each other. A projectile has both the horizontal o𝒓 𝒙 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 (𝑽𝒙) and vertical or y component (𝑽𝒚 ) components of motion. Each component is independent with each other.

Superposition of two motion: 1.) Constant velocity in the horizontal direction

𝑣𝑥 = 𝑣𝑖𝑥 = 𝑣𝑓𝑥 𝑎𝑥 = 0

2.) Constant acceleration in the vertical direction 𝑎𝑦 = 𝑔 = −9.8 𝑚/𝑠2 3.) Initial Velocity

𝑣𝑖𝑥 = 𝑣𝑖 𝑐𝑜𝑠𝜃 𝑣𝑖𝑦 = 𝑣𝑖 𝑠𝑖𝑛𝜃

𝑣𝑖 = 𝑣𝑖𝑥 𝑖 + 𝑣𝑖𝑦 𝑗 4.) Final Velocity

𝑣𝑓𝑥 = 𝑣𝑖𝑥

𝑣𝑓𝑦 = 𝑣𝑖𝑦 + 𝑎𝑦 𝑡

𝑣𝑓 = 𝑣𝑓𝑥 𝑖 + 𝑣𝑓𝑦𝑗 5.) Maximum Height

- maximum value of the vertical distance attained by the projectile .

−𝑉𝑖 2 𝑠𝑖𝑛2 𝜃 𝐻 = 𝑦𝑓 = 2𝑎𝑦

6.) Time to reach maximum height 𝑡=

−𝑣𝑖 𝑠𝑖𝑛𝜃 𝑎𝑦

7.) Horizontal Range or Horizontal Displacement - the distance from the point of projection to the point where the projection strikes the point in a horizontal plane. −𝑉𝑖 2 𝑠𝑖𝑛2𝜃 𝑅 = 𝑥𝑓 = 𝑎𝑦 8.) Time of flight – the time from the instant when it is projected to the time when it strikes the point in a horizontal plane passing through the point.

𝑇=

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−2𝑉𝑖 𝑠𝑖𝑛𝜃 𝑎𝑦

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1. To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 300 m/s at 55.0°above the horizontal. It explodes on the mountainside 42.0s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?

3. A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. a.) Determine the time of flight, b.) the horizontal distance, c.) the peak height of the longjumper, d.) the time it will reach the peak height, e.) the x and y component at t=0.80s and f.) express the velocity at t=0.80s in unit vector and determine its magnitude.

a.)Time of Flight

2. A stone is thrown from the top of a building upward at an angle of 30.0° to the horizontal with an initial speed of 20.0 m/s as shown in figure. The height of the building is 45.0 m.

b.) Horizontal Distance (Range)

c.) Peak Height (maximum height)

a.) How long does it take the stone to reach the ground?

d.) Time to reach maximum height

e.) the x and y component at t=0.80s b.) What is the speed of the stone just before it strikes the ground?

f.) vector and magnitude at t=0.80s

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