● Learn the concepts of displacement, velocity, and accelerationin one-dimension.

● Describe motions at constant acceleration.

● Be able to graph and interpret graphs as they describe motion.

● Be able to reason proportionally.

● Examine the special case of freely falling bodies.

● Consider relative motion.

Chapter 2 Motion Along a Straight Line(Motion in One-Dimension)

The study of motion is divided into two areas:

● Kinematics

● This will be our focus in Chapters 2 (1-dimension) and 3 (2-dimensions).

● Kinematics describes the movement of the object, regarding its position,

velocity, and acceleration.

●A coordinate system needs to be set up in order to describe its position,

velocity, and acceleration.

● Dynamics

●Will come in Chapter 4 and after

● Dynamics answers the question "Why is this object moving?“

● It is about how the application of forces affects the motion of an object.

2.1 Displacement and Average Velocity

● Coordinate system in 1-dimension: a single axis with a direction and an origin

Example: x-axis or y-axis

● Displacement describes the change in position:

● It is a vector quantity pointing from the initial position (x1) to the final position (x2).

● Its magnitude (the distance) is equal to the length of the straight line connecting these

two positions.

● In 1-dimension, the direction of a vector is either + or -, depending on if it is pointing in

the positive direction of the axis or not.

● Units in SI unit system: meters (m)

●Average velocity: The displacement divided by the elapsed time Dt = t2 – t1

● It is also a vector. Direction in 1-dimension: either + or -.

Displacement and Average Velocity

Displacement and Average Velocity

Understand the Meaning of Average Velocity Graphically

The average velocity between two positions is the slope of the line connecting these

two corresponding points on a graph of position as a function of time.

2.2 Instantaneous Velocity

Instantaneous Velocity:

How do we calculate the velocity at an instant of time?

Understanding the direction and the magnitude of the instantaneous velocity graphically

2.3 Average and Instantaneous Acceleration

When velocity changes with time acceleration

Units: m/s2

Average Acceleration

𝑎𝑎𝑣,𝑥 =𝑣2𝑥 − 𝑣1𝑥𝑡2 − 𝑡1

=∆𝑣𝑥∆𝑡

Instantaneous Acceleration

𝑎𝑥 = lim∆𝑡→0

∆𝑣𝑥∆𝑡

Understand the Direction of Acceleration

Direction of Velocity and Acceleration in a Free Fall Problem

Free fall from rest Tossing a Ball upwardCoordinate

y

0ground

the way downv is negative

a = - 9.8 m/s2

the way up

v is positive

a = - 9.8 m/s2

the way down

v is negative

a = - 9.8 m/s2

at the top

v = 0

a = - 9.8 m/s2

at the top

v = 0

a = - 9.8 m/s2

Example 2.4 Average and instantaneous acceleration, p. 37

The velocity of the object is given: vx(t) = (60.0 m/s) + (0.500 m/s3)t2

(a) Find the change in velocity in the time interval between t1 = 1.00 s and t2 = 3.00 s.

Dvx = vx(t2) – vx(t1) = vx(3.00 s) – vx(1.00 s)

(b) Find the average acceleration during this time interval.

aav,x = Dvx/Dt, where Dt = t2 – t1 = (3.00 s) – (1.00 s) = 2.00 s

(c) Estimate the instantaneous acceleration at time t1 = 1.00 s by taking Dt = 0.10 s.

aav,x = Dvx/Dt = [vx(1.10 s) – vx(1.00 s)]/(0.10 s)

2.4 Motion with Constant Acceleration

● Main goal: Understand the three (+1) kinematic equations and use them to solve problems

Question: How can we predict the motion of a particle?

●We must at least know where it starts at (x0) and what its initial velocity (v0x) is.

● These are called the initial conditions (at time t = 0, or, any given time) in a given

coordinate (x-axis).

Now Consider: a particle that is moving with a constant acceleration, ax = constant.

Question: What is the velocity of the particle at any other time t?

● Since the acceleration is defined by 𝑎𝑥 = 𝑣𝑥 𝑡 −𝑣𝑥(0)

𝑡−0= 𝑣𝑥 𝑡 −𝑣0𝑥

𝑡,

the velocity of the particle a time t is 𝑣𝑥 𝑡 = 𝑣0𝑥 + 𝑎𝑥𝑡 ………......(2.6)

● This is the first of the three kinematic equations.

Question: How can we predict the position of the particle at any time t?

● Since the average velocity is defined by 𝑣𝑎𝑣,𝑥 =𝑥 𝑡 − 𝑥(0)

𝑡 − 0=

𝑥 𝑡 − 𝑥0

𝑡,

we have 𝑥 𝑡 = 𝑥0 + 𝑣𝑎𝑣,𝑥𝑡.● But, what is 𝑣𝑎𝑣,𝑥?

Since 𝑣𝑥 𝑡 is linear in time t, 𝑣𝑥 𝑡 = 𝑣0𝑥 + 𝑎𝑥𝑡 ,

we have 𝑣𝑎𝑣,𝑥 =1

2[𝑣𝑥 𝑡 + 𝑣0𝑥] .

Therefore,

𝑥 𝑡 = 𝑥0 +1

2[𝑣𝑥 𝑡 + 𝑣0𝑥]𝑡

= 𝑥0 +1

2[ 𝑣0𝑥 + 𝑎𝑥𝑡 + 𝑣0𝑥]𝑡 ,

or, 𝑥 𝑡 = 𝑥0 + 𝑣0𝑥𝑡 +1

2𝑎𝑥𝑡

2 ……(2.10)

● This is the second of the three kinematic equations.

Question: What if you encounter a situation in which time is neither given nor asked of?

● Here comes the third of the three kinematic equations, which relates to each other the

position, velocity, and acceleration of a particle.

Let us consider the first two kinematic equations:

𝑣𝑥 𝑡 = 𝑣0𝑥 + 𝑎𝑥𝑡 ………………..…......(2.6)

𝑥 𝑡 = 𝑥0 + 𝑣0𝑥𝑡 +1

2𝑎𝑥𝑡

2 ……………..…(2.10)

The goal is to use substitution to eliminate time t.

Solve for time in terms of other quantities using (2.6): 𝑡 =𝑣𝑥 − 𝑣0𝑥

𝑎𝑥,

and then substitute into (2.10) to eliminate time:

𝑥 = 𝑥0 + 𝑣0𝑥𝑣𝑥 − 𝑣0𝑥

𝑎𝑥+

𝑎𝑥

2

𝑣𝑥 − 𝑣0𝑥

𝑎𝑥

2

= 𝑥0 +𝑣𝑥2−𝑣0𝑥

2

2𝑎𝑥,

or, 𝑣𝑥2 = 𝑣0𝑥

2 + 2𝑎𝑥(𝑥 − 𝑥0) …………………(2.11)

● This is the third of the kinematic equations.

Application of Kinematic Equations 𝑣𝑥 𝑡 = 𝑣0𝑥 + 𝑎𝑥𝑡 ………......(2.6)

𝑥 𝑡 = 𝑥0 + 𝑣0𝑥𝑡 +1

2𝑎𝑥𝑡

2……(2.10)

𝑣𝑥2 = 𝑣0𝑥

2 + 2𝑎𝑥 𝑥 − 𝑥0 … …(2.11)

𝑣𝑎𝑣,𝑥 =1

2[𝑣𝑥 𝑡 + 𝑣0𝑥]…..……(2.7)

Example 2.9 on page 45

(a) Find the time it takes for the officer to catch up with

the motorist? The approach: Find the expressions as

functions of time for the positions of the police and the

motorist, and then, set these two to equal to each other

to solve for time.

Position of the Officer: 𝑥𝑃 𝑡 = 0 + 0 +𝑎𝑃

2𝑡2

Position of the motorist: 𝑥𝐶 𝑡 = 0 + 𝑣𝐶0𝑡

Let 𝑥𝑃 𝑡 = 𝑥𝐶 𝑡 to solve for t:𝑎𝑃

2𝑡2 = 𝑣𝐶0𝑡,

which is the same as 𝑡(𝑎𝑃

2𝑡 − 𝑣𝐶0) = 0,

or 𝑡 =2𝑣𝐶0

𝑎𝑃= 10 s.

(b) What is the speed of the

officer at that time?

𝑣𝑃 𝑡 = 0 + 𝑎𝑃 = 𝑎𝑃2𝑣𝐶0𝑎𝑃

= 2𝑣𝐶0 = 30 m/s

(c) The distance the office has

traveled at that point,

𝑥𝑃 𝑡 = 0 + 0 +𝑎𝑃2𝑡2

=𝑎𝑃

2(2𝑣𝐶0

𝑎𝑃)2

=2𝑣𝐶0

2

𝑎𝑃= 150 m

2.5 Proportional Reasoning

Pay attention to the functional dependence, rather than focusing on numerical numbers.

An object, initially at rest, starts to

accelerate at t = 0 with a constant

acceleration. If it gains a speed 10

m/s after a certain time, what is its

speed when the time is doubled?

Equation to consider:

𝑣𝑥 𝑡 = 𝑣0𝑥 + 𝑎𝑥𝑡 = 𝑎𝑥𝑡with 𝑣0𝑥 = 0.

An object, initially at rest, starts to

accelerate at t = 0 with a constant

acceleration. If it travels a distance

10 m in the first 5 s, what is the

distance it travels in the first 10 s?

Equation to consider:

𝑥 𝑡 = 𝑥0 + 𝑣0𝑥𝑡 +1

2𝑎𝑥𝑡

2

with 𝑥0 = 0 and 𝑣0𝑥 = 0.

If the gravitational attraction a star

exerts on a planet is F when their

distance is R, what should be the

gravitational attraction if their

distance is doubled?

Equation to consider:

𝐹 = 𝐺𝑀𝑚

𝑅2

If the gravitational potential energy

between a star and a planet is V

when their distance is R, what

should be the gravitational potential

energy if their distance is cut in

half?

Equation to consider:

𝑉 = − 𝐺𝑀𝑚

𝑅

2.6 Freely Falling Objects

𝑣𝑦 𝑡 = 𝑣0𝑦 + 𝑎𝑦𝑡 ……….....(2.6)

y 𝑡 = 𝑦0 + 𝑣0𝑦𝑡 +1

2𝑎𝑦𝑡

2 . .…(2.10)

𝑣𝑦2 = 𝑣0𝑦

2 + 2𝑎𝑦 𝑦 − 𝑦0 ……(2.11)

𝑣𝑎𝑣,𝑦 =1

2[𝑣𝑦 𝑡 + 𝑣0𝑦] ………(2.7)

𝑎𝑦 = −𝑔 = − 9.8 𝑚/𝑠2 if the y-axis points up

𝑣𝑦 𝑡 = 𝑣0𝑦 − 𝑔𝑡 …………....(2.6)

y 𝑡 = 𝑦0 + 𝑣0𝑦𝑡 −1

2𝑔𝑡2 ……(2.10)

𝑣𝑦2 = 𝑣0𝑦

2 − 2𝑔 𝑦 − 𝑦0 ….….(2.11)

𝑣𝑎𝑣,𝑦 =1

2[𝑣𝑦 𝑡 + 𝑣0𝑦]….….…(2.7)

Example 2.10 on page 48

With the given y-axis and initial conditions, calculate the

position using eqn.(2.10) and the velocity using eqn.(2.6).

Pay attention to the sign of each quantity.

𝑣𝑦 𝑡 = 𝑣0𝑦 − 𝑔𝑡 …………......(2.6)

y 𝑡 = 𝑦0 + 𝑣0𝑦𝑡 −1

2𝑔𝑡2 ….…(2.10)

𝑣𝑦2 = 𝑣0𝑦

2 − 2𝑔 𝑦 − 𝑦0 …...….(2.11)

𝑣𝑎𝑣,𝑦 =1

2[𝑣𝑦 𝑡 + 𝑣0𝑦]…….…..(2.7)

Example 2.11 A ball on the roof on p. 11-12

Given: y0 = 0 and v0y = 15.0 m/s

Find: (a) y and vy at t = 1.00 s, and, y and vy at t = 4.00 s

(b) vy when y = 5.00 m

(c) maximum height ymax and the time reaching ymax

(a) Use eqn.(2.10) for y and use eqn.(2.6) for vy.

(b) There are more than one ways for calculating vy at y = 5.00 m.

(b1) Use eqn.(2.11), with v0y = 15.0 m/s and (𝑦 − 𝑦0) = 5.00 m.

You get two answers when you take the square-root:

vy = ± 11.3 m/s,

corresponding to the velocities on the ways up and down.

(b2) A less convenient way is to use eqn.(2.10) to solve for the

times at which the ball crosses y = 5.00 m, on the ways up

and down.

Then, use eqn.(2.6) to calculate the corresponding

velocities.

𝑣𝑦 𝑡 = 𝑣0𝑦 − 𝑔𝑡 ……….…....(2.6)

y 𝑡 = 𝑦0 + 𝑣0𝑦𝑡 −1

2𝑔𝑡2 ……(2.10)

𝑣𝑦2 = 𝑣0𝑦

2 − 2𝑔 𝑦 − 𝑦0 …..….(2.11)

𝑣𝑎𝑣,𝑦 =1

2[𝑣𝑦 𝑡 + 𝑣0𝑦]………..(2.7)

Example 2.11 A ball on the roof on p. 11-12

Given: y0 = 0 and v0y = 15.0 m/s

Find: (a) y and vy at t = 1.00 s, and, y and vy at t = 4.00 s

(b) vy when y = 5.00 m

(c) maximum height ymax and the time reaching ymax

(c) Meaning of “at the maximum height”? vy = 0.

So, you are asked to find the ymax and tmax for which vy = 0.

There are more than one ways.

(c1) Use eqn.(2.11) to find ymax, with y0 = 0, vy = 0, and

v0y = 15.0 m/s. Then, use eqn.(2.10) to find the time tmax

that it reaches the top.

(c2) use eqn.(2.6) to find the time tmax that it reaches the top.

Then, use eqn.(2.10) to calculate ymax.

(c3) Once you got tmax, use eqn.(2.7) to calculate ymax

ymax = 𝑣𝑎𝑣,𝑦tmax = [(𝑣𝑦 𝑡 + 𝑣0𝑦)/2] tmax

2.7 Relative Velocity Along a Straight Line (in One-Dimension)

Positions are relative: 𝑥𝑊/𝐶 = 𝑥𝑊/𝑇 + 𝑥𝑇/𝐶

Velocities are also relative: 𝑣𝑊/𝐶 = 𝑣𝑊/𝑇 + 𝑣𝑇/𝐶

Pay attention to the sign: position and velocity are vectors.