Chapter 2 Motion in One Dimension Kinematics n

Chapter 2 Motion in One Dimension

Kinematics n n Describes motion while ignoring the agents that caused the motion For now, will consider motion in one dimension n n Along a straight line Will use the particle model n A particle is a point-like object, has mass but infinitesimal size

Position n Defined in terms of a frame of reference n n One dimensional, so generally the x- or yaxis The object’s position is its location with respect to the frame of reference

Position-Time Graph n n The position-time graph shows the motion of the particle (car) The smooth curve is a guess as to what happened between the data points

Displacement n Defined as the change in position during some time interval n n n Represented as x x = xf - x i SI units are meters (m) x can be positive or negative Different than distance – the length of a path followed by a particle

Vectors and Scalars n Vector quantities need both magnitude (size or numerical value) and direction to completely describe them n n Will use + and – signs to indicate vector directions Scalar quantities are completely described by magnitude only

Average Velocity n n The average velocity is rate at which the displacement occurs The dimensions are length / time [L/T] The SI units are m/s Is also the slope of the line in the position – time graph

Average Speed n Speed is a scalar quantity n n n same units as velocity total distance / total time The average speed is not (necessarily) the magnitude of the average velocity

Instantaneous Velocity n n 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

Instantaneous Velocity n n The general equation for instantaneous velocity is The instantaneous velocity can be positive, negative, or zero

Instantaneous Velocity n n n The instantaneous velocity is the slope of the line tangent to the x vs. t curve This would be the green line The blue lines show that as t gets smaller, they approach the green line

Instantaneous Speed n n The instantaneous speed is the magnitude of the instantaneous velocity Remember that the average speed is not the magnitude of the average velocity

Average Acceleration n Acceleration is the rate of change of the velocity Dimensions are L/T 2 SI units are m/s²

Instantaneous Acceleration n The instantaneous acceleration is the limit of the average acceleration as t approaches 0

Instantaneous Acceleration n The slope of the velocity vs. time graph is the acceleration The green line green represents the instantaneous acceleration The blue line is the blue average acceleration

Acceleration and Velocity n n When an object’s velocity and acceleration are in the same direction, the object is speeding up When an object’s velocity and acceleration are in the opposite direction, the object is slowing down

Acceleration and Velocity n n The car is moving with constant positive velocity (shown by red arrows maintaining the same size) Acceleration equals zero

Acceleration and Velocity n n Velocity and acceleration are in the same direction Acceleration is uniform (blue arrows maintain the same length) Velocity is increasing (red arrows are getting longer) This shows positive acceleration and positive velocity

Acceleration and Velocity n n Acceleration and velocity are in opposite directions Acceleration is uniform (blue arrows maintain the same length) Velocity is decreasing (red arrows are getting shorter) Positive velocity and negative acceleration

1 D motion with constant acceleration tf – ti = t

1 D motion with constant acceleration n In a similar manner we can rewrite equation for average velocity: n and than solve it for xf n Rearranging, and assuming

1 D motion with constant acceleration (1) Using (1) and than substituting into equation for (2) final position yields (2) Equations (1) and (2) are the basic kinematics equations

1 D motion with constant acceleration These two equations can be combined to yield additional equations. We can eliminate t to obtain Second, we can eliminate the acceleration a to produce an equation in which acceleration does not appear:

Kinematics with constant acceleration - Summary

Kinematics - Example 1 n How long does it take for a train to come to rest if it decelerates at 2. 0 m/s 2 from an initial velocity of 60 km/h? 60 km/h

Kinematics - Example 1 n How long does it take for a train to come to rest if it decelerates at 2. 0 m/s 2 from an initial velocity of 60 km/h? 60 km/h n Using we rearrange to solve for t: n Vf = 0. 0 km/h, vi=60 km/h and a= -2. 0 m/s 2.

Kinematic Equations -- summary

Kinematic Equations n n n The kinematic equations may be used to solve any problem involving onedimensional motion with a constant acceleration You may need to use two of the equations to solve one problem Many times there is more than one way to solve a problem

A car is approaching a hill at 30. 0 m/s when its engine suddenly fails just at the bottom of the hill. The car moves with a constant acceleration of – 2. 00 m/s 2 while coasting up the hill. (a) Write equations for the position along the slope and for the velocity as functions of time, taking x = 0 at the bottom of the hill, where vi = 30. 0 m/s. (b) Determine the maximum distance the car rolls up the hill.

(a) Take at the bottom of the hill where xi=0, =0 vi=30 m/s, =30 m/s a=-2 m/s 2. Use these values in the general equation

(a) Take at the bottom of the hill where xi=0, vi=30 m/s, a=-2 m/s 2. Use these values in the general equation

The distance of travel, xf, becomes a maximum, when (turning point in the motion). Use the expressions found in part (a) for when t=15 sec.

Graphical Look at Motion: displacement-time curve n n The slope of the curve is the velocity The curved line indicates the velocity is changing n Therefore, there is an acceleration

Graphical Look at Motion: velocity-time curve n n The slope gives the acceleration The straight line indicates a constant acceleration

Graphical Look at Motion: acceleration-time curve n The zero slope indicates a constant acceleration

Freely Falling Objects n n A freely falling object is any object moving freely under the influence of gravity alone. It does not depend upon the initial motion of the object n n n Dropped – released from rest Thrown downward Thrown upward

Acceleration of Freely Falling Object n n The acceleration of an object in free fall is directed downward, regardless of the initial motion The magnitude of free fall acceleration is g = 9. 80 m/s 2 n n n g decreases with increasing altitude g varies with latitude g 9. 80 m/s 2 is the average at the Earth’s surface

Acceleration of Free Fall n n We will neglect air resistance Free fall motion is constantly accelerated motion in one dimension Let upward be positive Use the kinematic equations with ay = g = -9. 80 m/s 2

Free Fall Example n n n Initial velocity at A is upward (+) and acceleration is g (-9. 8 m/s 2) At B, the velocity is 0 and the 0 acceleration is g ( g -9. 8 m/s 2) At C, the velocity has the same magnitude as at A, but is in the opposite direction

A student throws a set of keys vertically upward to her sorority sister, who is in a window 4. 00 m above. The keys are caught 1. 50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught?

A student throws a set of keys vertically upward to her sorority sister, who is in a window 4. 00 m above. The keys are caught 1. 50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? (a)

A student throws a set of keys vertically upward to her sorority sister, who is in a window 4. 00 m above. The keys are caught 1. 50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? (b)

A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.

A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. 1 -st ball: 2 -nd ball:

A freely falling object requires 1. 50 s to travel the last 30. 0 m before it hits the ground. From what height above the ground did it fall?

A freely falling object requires 1. 50 s to travel the last 30. 0 m before it hits the ground. From what height above the ground did it fall? Consider the last 30 m of fall. We find its speed 30 m above 30 m the ground:

A freely falling object requires 1. 50 s to travel the last 30. 0 m before it hits the ground. From what height above the ground did it fall? Now consider the portion of its fall above the 30 m point. We 30 m assume it starts from rest Its original height was then:

Motion Equations from Calculus n n Displacement equals the area under the velocity – time curve The limit of the sum is a definite integral

Kinematic Equations – General Calculus Form

Kinematic Equations – Calculus Form with Constant Acceleration n The integration form of vf – vi gives n The integration form of xf – xi gives

The height of a helicopter above the ground is given by h = 3. 00 t 3, where h is in meters and t is in seconds. After 2. 00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

The height of a helicopter above the ground is given by h = 3. 00 t 3, where h is in meters and t is in seconds. After 2. 00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

The height of a helicopter above the ground is given by h = 3. 00 t 3, where h is in meters and t is in seconds. After 2. 00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? The equation of motion of the mailbag is:

Automotive engineers refer to the time rate of change of acceleration as the "jerk. " If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, speed, and position are axi , vxi, and xi , respectively. (b) Show that

Automotive engineers refer to the time rate of change of acceleration as the "jerk. " If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, speed, and position are axi , vxi, and xi , respectively. (b) Show that (a) constant when

Automotive engineers refer to the time rate of change of acceleration as the " jerk. " If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, speed, and position are axi , vxi, and xi , respectively. (b) Show that (a) when

Automotive engineers refer to the time rate of change of acceleration as the " jerk. " If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, speed, and position are axi , vxi, and xi , respectively. (b) Show that (a) when

(b) Recall the expression for v :

The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by a = – 3. 00 v 2 for v > 0. If the marble enters this fluid with a speed of 1. 50 m/s, how long will it take before the marble's speed is reduced to half of its initial value?

The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by a = – 3. 00 v 2 for v > 0. If the marble enters this fluid with a speed of 1. 50 m/s, how long will it take before the marble's speed is reduced to half of its initial value?

The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by a = – 3. 00 v 2 for v > 0. If the marble enters this fluid with a speed of 1. 50 m/s, how long will it take before the marble's speed is reduced to half of its initial value?