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Chapter 2: Motion in One Dimension: 2.1 Position, Velocity, and Speed

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Chapter 2: Motion in One Dimension

2.1 Position, Velocity, and Speed

The particle model: we describe the moving object as a particle regardless


of its size

A particle’s position is the location of the particle with respect to a chosen


reference point that we can consider to be the origin of a coordinate system.
position–time graph

The displacement of a particle is defined as its change in position in some


time interval.

Distance is the length of a path followed by a particle

In physics the terms speed and velocity are not the same

The average velocity = ∆x / ∆t

The average speed = total distance / total time

Displacement and velocity are vector quantities (require the specification of


both direction and magnitude).
By contrast, a scalar quantity has a numerical value and no direction.
Example 2.1

Find the displacement, average velocity, and average speed of the car in the
previous figure between positions A and F. (assume that the details of the
car’s position are described by the previous curve)

Average velocity= -1.7 m/s

Average speed = 2.5 m/s

2.2 Instantaneous Velocity and Speed

We need to know the velocity of a particle at a particular instant in time,


rather than the average velocity over a finite time interval.

The instantaneous velocity ʋx equals the limiting value of the ratio ∆x/∆t as
∆t approaches zero

the derivative of x with respect to t


The instantaneous speed of a particle is defined as the magnitude of its
instantaneous velocity

From here on, velocity means instantaneous velocity and speed means
instantaneous speed.
Example 2.2

Consider the following one-dimensional motions: (A) A ball thrown directly


upward rises to a highest point and falls back into the thrower’s hand. (B) A
race car starts from rest and speeds up to 100 m/s. (C) A spacecraft drifts
through space at constant velocity. Are there any points in the motion of
these objects at which the instantaneous velocity has the same value as the
average velocity over the entire motion?
If so, identify the point(s).

See page 30

Example 2.3 (see page 30)

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. The
position–time graph for this motion is shown in the following figure.

(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.
(B) Calculate the average velocity during these two time intervals.
(C) Find the instantaneous velocity of the particle at t = 2.5 s

(Note 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 (from a previous exam)

The motion of two cars A and B moving in a straight line is described by:

xA(t)= 4t2 and xB(t)= 6-5t+5t2

the two cars will reach the same velocity at a time (in seconds):

(a) 5.4 s (b) 1 s (c) 1.3 s (d) 0.28 s (e) 2.5 s


2.3 Acceleration

The average acceleration

The instantaneous acceleration

The acceleration can be positive or negative (vector)


Negative acceleration does not necessarily mean that an object is slowing
down.

When the object’s velocity and acceleration are in the same direction, the
object is speeding up. On the other hand, when the object’s velocity and
acceleration are in opposite directions, the object is slowing down.
Example 2.5 see page 33

The velocity of a particle moving along the x axis varies in time according to
the expression v=(40-5t2) m/s, where t is in seconds.
(a) Find the average acceleration in the time interval t =0 to t = 2.0 s.
(-10 m/s2)
(b) Determine the acceleration at t = 2.0 s.
-20 m/s2

2.5 One-Dimensional Motion with Constant Acceleration

The kinematic equations for motion of a particle under constant acceleration:

Try to prove these equations (in lecture)


Example 2.7
A jet lands on an aircraft carrier at 140 mi/h (~63 m/s).

(A)What is its acceleration (assumed constant) if it stops in 2.0 s.


(-31m/s2)
(B) If the plane touches down at position xi=0, what is the final position
of the plane?
63m

Example 2.8 see page 39


A car traveling at a constant speed of 45.0 m/s passes a trooper hidden
behind a billboard. One second after the speeding car passes the billboard,
the trooper sets out from the billboard to catch it, accelerating at a constant
rate of 3.00 m/s2. How long does it take her to overtake the car?
~31s
2.6 Freely Falling Objects

A freely falling object is any object moving freely under the influence of
gravity alone, regardless of its initial motion. (upward or downward)

Any freely falling object experiences an acceleration directed downward,


regardless of its initial motion.

In the absence of air resistance, all objects dropped near the Earth’s surface
fall toward the Earth with the same constant acceleration (g) under the
influence of the Earth’s gravity

At the Earth’s surface, the value of g is approximately 9.80 m/s2 (decreases


with increasing altitude)

for freely falling objects, the motion is in the vertical direction and the
acceleration is downward
ay = g = -9.80 m/s2.

Example 2.10 (p 41)

A ball is tossed straight up at 25 m/s. Estimate its velocity at 1 second


intervals.
Example

Which values represent the ball’s vertical velocity and acceleration at points
A, C and E in the following figure?

(a) vy= 0, ay = - 9.80 m/s2


(b) vy= 0, ay = 9.80 m/s2
(c) vy= 0, ay= 0
(d) vy = - 9.80 m/s, ay = 0

Example 2.12 (p 43)


A stone thrown from the top of a building is given an initial velocity of 20
m/s straight upward. The building is 50 m high, and the stone just misses the
edge of the roof on its way down. Using tA = 0 as the time the stone leaves
the thrower’s hand at position A, determine:

(A) The time at which the stone reaches its maximum height
(B) The maximum height,
(C) The time at which the stone returns to the height from which it was
thrown,
(D) The velocity of the stone at this instant
(E) The velocity and position of the stone at t = 5 s.
For illustration, see the following figure

Try to solve these problems (4-6-16-21-25-29-33-40-42-48-52) from chapter


two (p50-54) (some of them will be solved in lectures)

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