Ch. 2: Describing Motion: Kinematics in One Dimension

Terminology • Classical Mechanics = The study of objects in motion.

– 2 parts to mechanics.

• Kinematics = A description of HOW objects move. – Chapters 2 & 4 (Ch. 3 is mostly math!)

• Dynamics = WHY objects move.– Introduction of the concept of FORCE.– Causes of motion, Newton’s Laws– Most of the course after Chapter 4– For a while, assume ideal point masses (no– physical size). – Later, extended objects with size.

A Brief Overview of the Course “Point” Particles & Large Masses

• Translational Motion = Straight line motion. – Chapters 2,3,4,5,6,7,8,9

• Rotational Motion = Moving (rotating) in a circle.– Chapters 5,6,10,11

• Oscillations = Moving (vibrating) back & forth in same path.– Chapter 15

Continuous Media• Waves, Sound

– Chapters 15,16

• Fluids = Liquids & Gases– Chapter 13

Conservation Laws: Energy, Momentum, Angular Momentum– Just Newton’s Laws expressed in other forms!

THE COURSE THEME IS

NEWTON’S LAWS OF MOTION!!

Chapter 2 Topics

•Reference Frames & Displacement

•Average Velocity

•Instantaneous Velocity

•Acceleration

•Motion at Constant Acceleration

•Solving Problems

•Freely Falling Objects

Reference Frames & Displacement • Every measurement must be made with respect to a reference

frame. Usually, the speed is relative to the Earth.

• For example, if you are sitting on a train & someone walks down the aisle, the person’s speed with respect to the train is a few km/hr, at most. The person’s speed with respect to the ground is much higher.

• Specifically, if a person walks towards the front of a train at 5 km/h (with respect to the train floor) & the train is moving 80 km/h with respect to the ground. The person’s speed, relative to the ground is 85 km/h.

Coordinate Axes • Usually, we define a reference frame using a

standard coordinate axes. (But the choice of reference frame is arbitrary & up to us, as we’ll see later!)

• 2 Dimensions (x,y)

• Note, if its convenient, we could reverse + & - !

+,+- ,+

- , - + , -A standard set of xy (Cartesian or rectangular)

coordinate axes

Coordinate Axes • In 3 Dimensions (x,y,z)

• We define directions using these.

First Octant

Displacement & Distance The Distance traveled by an object

The Displacement of the object!

• Displacement The change in position of an object.

• Displacement is a vector (magnitude & direction).

• Distance is a scalar (magnitude).

In this figure, the

Distance = 100 m. The

Displacement = 40 m East.

Displacement

x1 = 10 m, x2 = 30 m

Displacement ∆x = x2 - x1 = 20 m

∆ Greek letter “delta” meaning “change in”

t1

t2

times

x1 = 30 m, x2 = 10 mDisplacement ∆x = x2 - x1 = - 20 m

Displacement is a VECTOR

Vectors and Scalars • Many quantities in physics, like displacement,

have a magnitude and a direction. Such quantities are called VECTORS.– Other quantities which are vectors: velocity,

acceleration, force, momentum, ...• Many quantities in physics, like distance, have

a magnitude only. Such quantities are called SCALARS.– Other quantities which are scalars: speed,

temperature, mass, volume, ...

• The Text uses BOLD letters to denote vectors.

• I usually denote vectors with arrows over the symbol.

• In one dimension, we can drop the arrow and remember that a + sign means the vector points to right & a minus sign means the vector points to left.

Average Velocity Average Speed (Distance traveled)/(Time taken)

Average Velocity (Displacement)/(Time taken)

• Velocity: Both magnitude & direction describing how fast an object is moving. Velocity is a VECTOR. (Similar to displacement).

• Speed: Magnitude only describing how fast an object is moving. A SCALAR. (Similar to distance).

Units: distance/time = m/s

A Scalar!

A Vector!

Average Velocity, Average Speed • Displacement from before. Walk for 70 s.

• Average Speed = (100 m)/(70 s) = 1.4 m/s

• Average velocity = (40 m)/(70 s) = 0.57 m/s

timest2

t1

Bar denotes average

• In general:

∆x = x2 - x1 = displacement∆t = t2 - t1 = elapsed timeAverage Velocity: (x2 - x1)/(t2 - t1)

Example A person runs from x1 = 50.0 m to x2 = 30.5 m in ∆t = 3.0 s. ∆x = -19.5 m.

Average Velocity (∆x)/(∆t)

= -(19.5 m)/(3.0 s) = -6.5 m/s

The negative sign indicates the negativex direction.

Instantaneous Velocity velocity at any instant of time

average velocity over an infinitesimally short time• Mathematically, the instantaneous velocity is

defined by:

ratio considered as a whole for smaller & smaller ∆t. As you should know, mathematicians call this a derivative.

Instantaneous Velocity v ≡ Time Derivative of Displacement x

These graphs show (a) constant velocity

and

(b) varying velocity

instantaneous velocity = average velocity

instantaneous velocity average velocity

The instantaneous velocity is the average velocity in the limit as the time interval becomes infinitesimally short.

Ideally, a speedometer would measure instantaneous velocity; in fact, it measures average velocity, but over a very short time interval.

On a graph of a particle’s position vs. time, the instantaneous velocity is the slope of the tangent to the curve at any point.

A jet engine moves along an experimental track (called the x axis)

as shown. Treat the engine as if it were a particle. Its position as a function of time is given by the equation x = At2 + B, where A = 2.10 m/s2 & B = 2.80 m. (a) Find the displacement of the engine during the time interval from t1 = 3.0 s to t2 = 5.0 s. (b) Find the average velocity during this time interval. (c) Determine the magnitude of the instantaneous velocity at t = 5.0 s.

Example: Given x as a Function of t.

Work on the board!

Acceleration • Velocity can change with time. An object with

velocity that is changing with time is said to be accelerating.

• Definition: Average acceleration = ratio of change in velocity to elapsed time.

a = (v2 - v1)/(t2 - t1) – a

– Acceleration is a vector.

• Instantaneous acceleration

• Units: velocity/time = distance/(time)2 = m/s2

Example: Average Acceleration A car accelerates along a straight road from rest to 90 km/h in 5.0 s. Find the magnitude of its average acceleration. Note: 90 km/h = 25 m/s a =

A

Example: Average Acceleration A car accelerates along a straight road from rest to 90 km/h in 5.0 s. Find the magnitude of its average acceleration. Note: 90 km/h = 25 m/s a = = (25 m/s – 0 m/s)/5 s = 5 m/s2

Conceptual Question

1. Velocity & acceleration are both vectors.

Are the velocity and the acceleration

always in the same direction?

Conceptual Question 1. Velocity & acceleration are both vectors.

Are the velocity and the acceleration

always in the same direction?

NO!!

If the object is slowing down, the acceleration vector is in the opposite direction of the velocity vector!

Example: Car Slowing Down

A car moves to the right ona straight highway (positive

x-axis). The driver puts on the brakes. If the initial velocity (when the driver hits

the brakes) is v1 = 15.0 m/s.It takes 5.0 s to slow down to v2 = 5.0 m/s. Calculate the

car’s average acceleration. a =

Example: Car Slowing Down

a = = (v2 – v1)/(t2 – t1) = (5 m/s – 15 m/s)/(5s – 0s)

a = - 2.0 m/s2

A car moves to the right ona straight highway (positive

x-axis). The driver puts on the brakes. If the initial velocity (when the driver hits

the brakes) is v1 = 15.0 m/s.It takes 5.0 s to slow down to v2 = 5.0 m/s. Calculate the

car’s average acceleration.

Deceleration

• “Deceleration”: A word which means “slowing down”. We try to avoid using it in physics. Instead (in one dimension), we talk about positive & negative acceleration.

• This is because (for one dimensional motion) deceleration does not necessarily mean the acceleration is negative!

The same car is moving to the left instead of to the right. Still assume positive x is to the right. The car is decelerating & the initial & final velocities are the same as before. Calculate the average acceleration now.

Conceptual Question2. Velocity & acceleration are vectors.

Is it possible for an object to have a zero

acceleration and a non-zero velocity?

Conceptual Question2. Velocity & acceleration are vectors.

Is it possible for an object to have a zero

acceleration and a non-zero velocity?

YES!!

If the object is moving at a constant velocity, the acceleration vector is zero!

Conceptual Question3. Velocity & acceleration are vectors.

Is it possible for an object to have a zero

velocity and a non-zero acceleration?

Conceptual Question3. Velocity & acceleration are vectors.

Is it possible for an object to have a zero

velocity and a non-zero acceleration?

YES!!

If the object is instantaneously at rest (v = 0) but

is either on the verge of starting to move or is

turning around & changing direction, the

velocity is zero, but the acceleration is not!

One-Dimensional Kinematics Examples

As already noted, the instantaneous acceleration is the average acceleration in the limit as the time interval becomes infinitesimally short.

The instantaneous slope of the velocity versus

time curve is the instantaneous acceleration.

Example : Acceleration given x(t). A particle moves in a straight line so that its position is given by x = (2.10 m/s2)t2 + (2.80 m). Calculate: (a) its average acceleration during the time interval from t1 = 3 s to t2 = 5 s, & (b) its instantaneous acceleration as a function of

time.

position vs time curve velocity vs time curve acceleration vs time curve

Acceleration is constant here!

Velocity vs time curve is a straight line

In this case, position vs time curve is a parabola

Velocity v(t) = time derivative of position x(t): v(t) = = (4.2)t. Acceleration a(t) = time derivative of velocity v(t): a(t) = = 4.2 m/s2 . So, answer to part (b): 4.2 m/s2 (Note: Acceleration is the 2nd derivative of the position!) Solution to (a): At t1 = 3 s , v1 = (4.2)(3) = 12.6 m/s. At t2 = 5 s , v2 = (4.2)(5) = 21 m/s. So, average acceleration a = = (21 m/s – 12.6 m/s)/(5 s – 3 s) = 4.2 m/s2. It makes sense that the average & instantaneous accelerations are the same, because the acceleration is constant!

Example : Acceleration given x(t). A particle moves in a straight line so that its position is given by x = (2.10 m/s2)t2 + (2.80 m). Calculate: (a) its average acceleration during the time interval from t1 = 3 s to t2 = 5 s, & (b) its instantaneous acceleration as a function of

time. In this case, position vs time curve is a parabola

Velocity vs time curve is a straight line

Acceleration is constant here!

position vs time curve velocity vs time curve acceleration vs time curve

Conceptual Example: Analyzing with graphs: The figure shows the velocity v(t) as a function of time for 2 cars, both accelerating from 0 to 100 km/h in a time 10.0 s. Compare (a) the average acceleration; (b) the instantaneous acceleration; & (c) the total distance traveled for the 2 cars.

Solution: (a) Ave. acceleration: a = Both have the same ∆v & the same ∆t so a is the same for both.

(c) Total distance traveled: Except for t = 0 & t = 10 s, car A is moving faster than car B. So, car A will travel farther than car B in the same time.

Conceptual Example: Analyzing with graphs: The figure shows the velocity v(t) as a function of time for 2 cars, both accelerating from 0 to 100 km/h in a time 10.0 s. Compare (a) the average acceleration; (b) the instantaneous acceleration; & (c) the total distance traveled for the 2 cars.

(b) Instantaneous acceleration: a = = slope of tangent to v vs t curve. For about the first 4 s, curve A is steeper than curve B, so car A has greater a than car B for times t = 0 to t = 4 s. Curve B is steeper than curve A, so car B has greater a than car A for times t greater than about t = 4 s.

Example (for you to work!) : Calculating Average Velocity & Speed

Problem: Use the figure & table to find the displacement & the average velocity of the car between positions (A) & (F).

a) Calculate the displacement of the particle in the time intervals t = 0 to t = 1 s &

t = 1 s to t = 3 s (A to B & B to C)

b) Calculate the average velocity in the timeintervals t = 0 to t = 1 s & t = 1 s to t = 3 s

(A to B & B to C) c) Calculate the instantaneous velocity of the particle at t = 2.5 s (point C) .

Example (for you to work!) : Average & Instantaneous Velocity Problem: A particle moves along the x axis. Its x coordinate varies with time as x = -4t + 2t2. x is in meters & t is in seconds. The position-time graph for this motion is in the figure.

Example (for you to work!) : Graphical Relations between x, v, & a Problem: The position of an object moving along the x axis varies with time as in the figure. Graph the velocity versus time and acceleration versus time curves for the object.

Example (for you to work!) : Average & Instantaneous Acceleration

Problem: The velocity of a particle moving along the x axis varies in time according to the expression v = (44 - 10t 2), where t is in seconds.

a) Calculate the averageacceleration in the time interval t = 0 to t = 2.0 s.

b) Calculate the acceleration at t = 2.0 s.