Motion in One Dimension

Unit 1

Lesson 1 : Position, Velocity, and Speed

Position : location of a particle with respect to a chosen reference point

Displacement : the change in position in some time interval

x = xf – xi

Distance : the length of a path followed by a particle

Displacement is a VECTOR

QUANTITY

Distance is a SCALAR

QUANTITY

Average Velocity : a particle’s displacement (x) divided by the time interval (t)

during which that displacement occurs

V =x

tAverage Speed : the total distance traveled divided

by the total time interval required to travel that distance

Avg speed =total distance

total time

Avg Velocity is a VECTOR

QUANTITY

Avg Speed is a SCALAR

QUANTITY

Example 1

Position t(s) x(m)

A 0 30

B 10 52

C 20 38

D 30 0

E 40 -37

F 50 -53

Find the displacement, average velocity, and average speed of the object between

positions A and F.

Lesson 2 : Instantaneous Velocity and Speed

Instantaneous Velocity : the limiting value of the ratio x/t as t approaches zero

v =x

tlim

t 0

Instantaneous Speed : the magnitude of the instantaneous velocity

Example 1

A) A ball thrown directly upward rises to a highest point and falls back in 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 at which the instantaneous velocity has the same value as the average velocity over the entire motion ? If so, identify the point(s).

What is the instantaneous velocity at t = 5.0 s ?

Answer : Slope of the tangent line drawn at the time in question.

Example 2

Lesson 3 : AccelerationAverage Acceleration : the change in velocity

(v) divided by the time interval (t) during which that change occurs

a =v

t=

vf – vi

tf – ti

Instantaneous Acceleration : the limit of the average acceleration as t appoaches zero

a =v

tlim

t 0

Example 1

What is the instantaneous acceleration at t = 2.0 s ?

Answer : Slope of the tangent line drawn at the time in question.

Example 2

The velocity of a particle moving along the x-axis varies in time according to the expression

vx = (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.

B) Determine the acceleration at t = 2.0 s.

Lesson 4 : Motion Diagrams

v

No acc

Motion A

v

a

Motion B

v

a

Motion C

Graphs for each motion :

x

Motion A

t

v

t

at

x

Motion B

t

v

t

at

x

Motion C

t

v

t

at

Example 1

Draw the corresponding x vs. t and a vs. t graphs.

A)v

t

x

t

a

t

B)v

t

x

t

a

t

vC)

t

x

t

a

t

Example 2

Draw the corresponding v vs. t and a vs. t graphs.

A)x

t

v

t

a

t

B)x

t

v

t

a

t

xC)

t

v

t

a

t

What is the displacement of the object from t = 0 to 3 s ?

Answer : The area under the graph equals displacement

Example 3

Negative Area :

Object is moving toward smaller x

values and displacement is

decreasing

Lesson 5 : Kinematic Equations

a =v

t=

vf – vi

tf – ti

vf = vi + at

vf = vi + at (for constant a)

v =vi + vf

2(for constant a)

x = xf - xi

x = vt

xf – xi = vt

xf – xi = ½ (vi + vf)t

xf = xi + ½ (vi + vf)t (for constant a)

xf = xi + ½ (vi + vf)t

xf = xi + ½ [vi + (vi +at)]t

vf = vi + at

xf = xi + vit + ½ at2 (for constant a)

xf = xi + ½ (vi + vf)t

xf = xi + ½ (vi + vf) (vf - vi

a)

vf2 = vi

2 + 2a (xf – xi) (for constant a)

Summary

vf = vi + at (for constant a)

v =vi + vf

2(for constant a)

xf = xi + ½ (vi + vf)t (for constant a)

xf = xi + vit + ½ at2 (for constant a)

vf2 = vi

2 + 2a (xf – xi) (for constant a)

Example 1

A jet lands on an aircraft carrier at 140 mph (63 m/s).

A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the airplane and brings it to a stop ?

B) If the plane touches down at position xi = 0, what is the final position of the plane ?

Example 2

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 ?

Lesson 6 : Freely Falling Objects

Neglecting air resistance, all objects dropped near the Earth’s surface fall

toward the Earth with the same constant acceleration under the influence of the

Earth’s gravity.

A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial

motion.

Free-fall acceleration (g) = 9.80 m/s2

+y -y

ay = -g = -9.80 m/s2

For making quick estimates, use g = 10 m/s2

Example 1

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

Time (s) Velocity (m/s)

0 +25

1

2

3

4

5

Example 2

A stone is 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.

A) Determine the time at which the stone reaches its maximum height.

B) Determine the maximum height.

D) Determine the velocity of the stone at this instant.

E) Determine the velocity and position of the stone at t = 5.00 s.

C) Determine the time at which the stone returns to the height from which it was thrown.

Lesson 7 : Using Calculus in Kinematics

Instantaneous Velocity v =x

tlim

t 0

This limit is called the derivative of x with respect to t.

v =x

tlim

t 0

=dx

dt

v =dx

dt

f(x) = axn

Derivative of Power Function

dfdx

= naxn-1

Example 1

An object is moving in one dimension according to the formula x(t) = 2t3 + t2 – 4.

Find its velocity at t = 2 s.

Instantaneous Acceleration a =v

tlim

t 0

This limit is called the derivative of v with respect to t.

a =v

tlim

t 0

=dv

dt

a =dv

dt

Example 2

An object is moving in one dimension according to the formula v(t) = (40 – 5t2).

Find its acceleration at t = 2 s.

Acceleration is also the second derivative of position.

a =v

tlim

t 0

=dv

dt=

d2x

dt2

Example 3

An object is moving in one dimension according to the formula x(t) = 12 – 4t + 2t3.

Find its acceleration at t = 3 s.

The Integral or Antiderivative

v

tti tf

tn

vn

xn = vntn = shaded area

x = vntn = total area

As rectangles get narrower, tn 0

x = limtn 0

vn tn

Displacement = area under v-t graph

This is called the Definite Integral

limtn 0

vn tn = v(t) dtti

tf

Integrating Velocity

v(t) =dx

dt

dx = v dt

xi

xf

dx =ti

tf

v dt

xf – xi = ti

tf

v dt

Integrating Acceleration

a(t) =dv

dt

dv = a dt

vi

vf

dv =ti

tf

a dt

vf – vi = ti

tf

a dt

Finding Antiderivatives

If k is a constant,

k dx = kx + C

xn dx =xn+1

n + 1+ C (n = 1)

Example 4

A car currently moving at 10. m/s accelerates non-uniformly according to

a(t) = 3t2. Find its velocity at t = 2 s.

Example 5

An object is moving according to the formula v(t) = 5t2 – 4t. If the object starts

from rest, find its position at t = 5s .

Example 6

An object is moving according to the formula a(t) = 2t – 4, with an initial velocity of + 4 m/s, and an initial position x = 0 at

time t = 0. Find the position and velocity at arbitrary times.