kinematics in one dimension

Kinematics in One Dimension

Chapter 2

Expectations

After this chapter, students will:

distinguish between distance and displacement distinguish between speed and velocity understand what acceleration is recognize situations in which the kinematic

equations apply

Expectations

After this chapter, students will:

be able to do calculations involving freely-falling objects

Kinematics

Mechanics Kinematics: the descriptive study of motion,

without concern for what causes the motion (space and time only)

Dynamics: the study of what causes and changes the motion of objects (force and inertia)

Displacement and Distance

Displacement: the vector that extends from the starting point to the ending point of an object’s travel.

Distance: the total length of an object’s path (a scalar quantity).


An ant crawls along the edge of a meter stick.

He starts out at x0 = 50 cm

Then he moves along to x1 = 75 cm


Now he changes his mind and goes toward the left,

to x2 = 20 cm …

… and reverses course yet again and goes to

xf = 30 cm, where he collapses in weariness.


What distance did the ant travel? from x0 = 50 cm to x1 = 75 cm: d1 = 25 cm

from x1 = 75 cm to x2 = 20 cm: d2 = 55 cm

from x2 = 20 cm to xf = 30 cm: d3 = 10 cm

total distance = d1 + d2 + d3

= 25 cm + 55 cm + 10 cm = 90 cm


What was the ant’s displacement?

x = xf – x0 = 30 cm – 50 cm = - 20 cm

The magnitude of the displacement was 20 cm; the direction was toward – X (or, as we drew it, leftward).

ant’s initial position

ant’s fi nal position

X


Magnitude of displacement: x = xf – x0

Direction of displacement: from x0 to xf

SI unit: meter (m)

Speed and Velocity

Back to the ant-infested meter stick: but this time, we’ll take a stopwatch with us.

x0 = 50 cm

t0 = 0.0 s t1 = 5.0 s

x1 = 75 cm

Speed and Velocity

Note that speed is a scalar quantity.

Dimensions: length/time SI unit: m/s

x0 = 50 cm

t0 = 0.0 s t1 = 5.0 s

x1 = 75 cm

cm/s 5.0 s 5.0

cm 25

s 0.0 - s 5.0

cm 50 - cm 75 speed average

01

01

tt

xx

Speed and Velocity

Velocity is a vector: its direction is the same as that of the displacement vector.

Magnitude:

x0 = 50 cm

t0 = 0.0 s t1 = 5.0 s

x1 = 75 cm

ntdisplaceme 1

velocity average01

tt

cm/s 5.0 s 0.0 - s 0.5

cm 50 - cm 75

01

01

tt

xx

t

xv

Speed and Velocity

Now, the second part of the journey …

x2 = 20 cm

t2 = 19 s t1 = 5.0 s

x1 = 75 cm

Speed and Velocity

… and the third.

x2 = 20 cm

t2 = 19 s t f = 22 s

xf = 30 cm

Speed and Velocity

Times and positions:

Average speed:

t x

0.0 s 50 cm

5.0 s 75 cm

19 s 20 cm

22 s 30 cm

cm/s 4.1 s 0.0 - s 22

cm 10 cm 55 cm 25 speed ave.

speed ave.0

1

tt

xx

f

ii

Speed and Velocity

Times and positions:

Average velocity:

t x

0.0 s 50 cm

5.0 s 75 cm

19 s 20 cm

22 s 30 cm

left)(or X :direction

)(magnitude cm/s 0.91- s 0.0 - s 22

cm 50 - cm 30

0

0

v

tt

xx v

f

f

Acceleration

We’ve been talking about “average” velocity because velocity may not be constant as motion occurs.

If velocity changes, acceleration occurs.

Acceleration

Acceleration is the time rate of change of velocity.

If an object has a velocity v0 at time t0, and a different velocity v at a later time t, its acceleration is:

dimensions: length / time2 SI units: m/s2

0

0

tt

vva

Acceleration

Note that acceleration may or may not involve a change in speed. If either the magnitude (speed) or the direction (or both) of the velocity vector change, an acceleration has taken place.

In common speech, “acceleration” means that speed is increasing. In physics, acceleration means: speed increases; speed decreases; direction of motion changes; or some combination of these.

Acceleration

Consider an elevator car. At the initial time t0 = 0 s, it is descending with a velocity v0 = -1.2 m/s. It comes to a stop (v1 = 0 m/s) at its floor at time t1 = 1.5 s. The magnitude of its acceleration is

2

01

011 m/s 0.80

s 0 - s 1.5

m/s) (-1.2 - m/s 0

tt

vva v0 = -1.2 m/ s

v1 = 0 m/ sv1 – v0 = +1.2 m/ s

v0 (v1 – v0)

a1 = v1 – v0

t1 – t0

Acceleration

Passengers enter, the door closes, and the elevator starts upward, reaching a constant upward velocity of 1.7 m/s in a time of 3.0 s. The acceleration:

2

02

022 m/s 0.57

s 0 - s 3.0

m/s 0 - m/s .71

tt

vva

v0 = 0 m/ sv2 = +1.7 m/ s

v2 – v0 = +1.7 m/ s

v2 (v2 – v0)

a1 = v1 – v0

t1 – t0

Acceleration

The elevator slows as it approaches the 86th floor. At t3 = 177 s, its velocity (v3) is 1.7 m/s, upward; at t4 = 179.5 s, it stops (v4 = 0 m/s).

When the acceleration vector points in the opposite direction from the initial velocity vector, speed decreases.

2

34

344 m/s 0.68-

s 177 - s 179.5

m/s 1.7 - m/s 0

tt

vva

v4 = 0 m/ sv3 = +1.7 m/ s

v4 – v3 = -1.7 m/ s

v3 (v4 – v3)

a4 = v4 – v3

t4 – t3

Kinematic Equations

If acceleration is constant:

From the definition of acceleration:

atvv

ttt

ttavv

ttavvvtt

vva

0

0

00

000

0

:simply interval time thecall weIf

:for Solve

Kinematic Equations

If acceleration is constant, from the definition of average velocity:

tvvx

xvv

t

xv

0

0

2

1

:for Solve 2

Kinematic Equations

If acceleration is constant, we can substitute v from the first kinematic equation into the second:

20

000

00

2

1

22

1

2

1

2

1

attvx

tatvtatvvx

atvvtvvx

Kinematic Equations

If acceleration is constant, we can solve the definition of acceleration for t:

and substitute in the second kinematic equation:

a

vvt

t

vva 00

axvvva

vvx

a

vvvv

a

vvvvtvvx

2 :for Solve 2

22

1

2

1

20

222

02

00000

Kinematic Equations

Why do we insist that acceleration must be constant?

We are implicitly assuming that velocity is a linear function of time.

atvv

mxbybmxy

0

Kinematic Equations: Summary

Valid only when acceleration is constant.

axvv

attvx

tvvx

atvv

2

2

12

1

20

2

20

0

0

Graphical Representations of Motion

Object moving at constant velocity

t

x

t0 t

x0

xx-x0

t-t0

velocity = slope = x – x0

t – t0


Object moving with changing velocity

t

x

t0 t f

x0

xf

xf-x0

t f-t0

average velocity = xf – x0

t f – t0

instantaneous velocity = slope of tangent line at any point


Object moving with constant velocity

t

v

v

tt ft0

x = vt = area under velocity vs. time plot


Object moving with constant acceleration

t

v

tt ft0

v0

vf

slope = acceleration

x = v t = (v0 + v) t

(area under v vs. t plot)

12

not true if v is not a linear function of t (constant acceleration)

Free Fall ≈ Constant Acceleration

Objects at or near Earth’s surface experience a constant acceleration toward the Earth’s center, due to gravitation – if we ignore the resistive effect of the air through which it falls.

The magnitude of this acceleration is 9.807 m/s2.

The symbol of this acceleration is g.

Free Fall ≈ Constant Acceleration

In the presence of air, a falling object has an initial downward acceleration g; the acceleration decreases until the object is falling with a constant velocity called its terminal velocity.

The acceleration due to gravity, g, does not depend on the mass or the size of the object.

Free Fall and the Kinematic Equations

Consider a stone thrown straight upward with an initial velocity v0. It will ascend with its velocity decreasing until it stops momentarily; then it will accelerate back downward.

Let’s calculate the time, ta, of its ascent.

v0


First, we need to establish a (one-dimensional) coordinate system.

Our positive x-axis will point straight up.

That means the stone’s initial velocity is +v0, and its acceleration is –g.

v0

g

0

+X


In the first kinematic equation, we know everything except t. We know this because the final velocity, v, is zero (the stone stops momentarily at the top of its trajectory). Another way of looking at this is to say we know v = 0 when t = ta. We can substitute these into the first kinematic equation and solve it for ta.

v0

g

0

+Xatvv 0


v0

g

0

+X

g

vt

vgt

tgtv

ga

atvatvv

a

a

aa

a

0

0

0

00

:for Solve 0

0


Now, let’s calculate the height of the ascent.

We know that at the maximum height xmax, the stone stops momentarily; that is, v = 0 when x = xmax.

v0

g

0

+X


Substituting v = 0, a = -g, and x = xmax into our fourth kinematic equation: v0

g

0

+X

g

vxvgx

gxvaxvv

2 2

20 22

0max

20max

max2

02

02


As the stone falls back to the height (x = 0) from which it started, we wonder: how long does the descent take?

Will it take the same amount of time that the ascent did?

Let’s find out.

v0

g

0

+X


For the descent:

v0 = 0, x = -xmax, t = td, and a = -g.

Substitute these into the third kinematic equation:

v0

g

0

+X

g

xtxgt

gttxattvx

dd

dd

maxmax

2

2max

20

2

2

1

2

10-

2

1


That result doesn’t look much like our result for ta:

But, if we substitute our result for xmax into our equation for td:

ad tg

v

g

v

g

g

v

g

xt

g

vx

02

20

20

max

20

max

22

2

2

g

vta

0


By now, you probably suspect that the ascent and descent are fully symmetric, and would probably predict that the final velocity of the falling stone is equal in magnitude and opposite in direction to the initial velocity of the ascending stone.

We can check this using the first kinematic equation, substituting our calculated td for t, -g for a, zero for v0, and vf for v.

atvv 0


The result:

Thus, we have demonstrated that the ascent of an object thrown upward, and its descent to the same height, are indeed fully symmetric.

If the object returns to a different height … “all bets are off.”

0

00 0

vv

g

vgvatvv

f

f

kinematics in one dimension

Documents

magnitude speed

average speed

velocity changes

average velocity

journey speed

displacement vector

velocity v0

velocity vector change