uniform rectilinear motion uniform accelerated...

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UUnit 3: Kinematics

Uniform Rectilinear MotionUniform Accelerated Rectilinear MotionThe Motion of Projectiles

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p228229

We can use arrows to indicate direction and change of velocity along a straight line

The length of the arrow is proportional to the speed.

A positive acceleration and negative direction

Negative acceleration and positive direction

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graphs: slope of a line

slope = rise/run

if y vs x then slope = Δy/Δx

If the graph is displacement versus time:slope = Δd/Δt = d2 - d1/ t2 - t1 = velocity

If the graph is velocity versus time: slope = Δv/Δt = v2 - v1/ t2 - t1 = acceleration

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Uniform Motion:

Constant speed: an object moves equal distances in equal time intervals

Uniform Motion(: an object moves with constant velocity (constant speed and direction)

Graphs:

Velocity vs Time: uniform motion

v

t

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remember constant velocity on a displacement vs time graph?

d

t

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UARM

UNIFORMLY ACCELERATED RECTILINEAR MOTION

motion in a straight line accelerating at a constant rate

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constant acceleration

The car accelerates at 2 m/s2

means every second the car will go 2m/s faster

Time (s) Velocity (m/s)

0 0

1 2

2 4

3 6

4 8

5 10

6 12

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If a car is slowing down the accelertation will be 3m/s2

time (s) Velocity

0 30

1 27

2 24

3 21

4 18

5 15

6 12

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Aristotle

4th century BC

2 kinds of motion • violent (throw, push)• natural (fire rised, stones fall down)

Oresme

14 the century predicted that if the initial velocity was zero, distance was proportional to time squared

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Galileo

1564-1642

Did experiments:

marble on the inclined plane

he noticed a marble on an inclined plane took more timethan a marble dropped (free fall)

"the gravitational force has been diluted"

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450 years later

Apollo 15 lands on the moon

A feather falls at the same time as a hammer

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Acceleration

the change in velocity over time

a = Δv/Δt = m/s2

a truck's speed changes from 5 m/s to 50 m/s, in 60 seconds, what is its acceleration?

50m/s 5m/s / 60s = 45/60 m/s2

= .75m/s2

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Uniformly Accelerated Rectilinear Motion

velocity is no longer constant (more real)velocity varies moment to moment

ti = initial time (s)tf = final time (s)xi = initial position (m)xf = final position (m)vi = initial velocity (m/s)vf = final velcity (m/s)a = acceleration (m/s2)

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Instantaneous velocity

velocity at a precise moment in time

On a graph:position versus time, UARM

d

t

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take the tangent of the curve

d

t

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a = y2 - y1x2 - x1

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velocity vs time graphs

v

t

There is a constant change in velocity, this object is speeding up with uniform acceleration

a = Δv/Δt = ?

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velocity vs time graphs

v

t

Slope

Δv/Δt gives acceleration

AreaΔv*Δt = displacement

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Equations of Uniform Acceleration

Equation 1: a = Δv/Δt = v2 - v1/Δt a * Δt = v2 - v1 a * Δt + v1 = v2

v2 = v1 + at

Equation 3: xf =xi + v1t + ½at2

Equation 2: xf = xi + ½(vi + vf)Δt

Equation 4: vf2 = vi

2 + 2aΔx

Equation 1: v2 = v1 + at

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How to solve kinematic problems:textbox p232

• draw a diagram• origin of x axis, at starting point, xi = 0• id ti and tf

• id known parameters (be sure to indicate signs (+ or -)• Find one of the 4 equations where the quantity sought is the only unknown

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Example:

The driver of a car which is moving east at 25m/s applies the brakes and begins to decelerat at 2.0m/s2

How far does the car travel in 8.0s?

a = -2.0 m/s2

vi = 25 m/st = 8.0 s

d = ?

which formula?

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d = vit + 1/2 a t2

d = 25(8.0) + 1/2(-2.0)(8.02)d = 200 - 64d = 136 m (+)

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Examples in text book p 232

A:

B:

Practice: Section 10.2p. 234

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p. 2341. What do we know?

Δd = 402m vi= 0m/s Δt = 6.0s xi = 0m xf = 402m?a? vf (km/h) Look for formulas with only one unknown....

xf = xi + (viΔt + 1/2(aΔt2))

find a402 = 0 + (0*6 + 1/2(a*6^2)402 = 1/2(a*36)804 = 36aa = 22.3m/s2 = 22m/s2

vfvf2 = vi2 +2aΔxvf2 = 0 + 2* 22.3 * 402

vf = √17929.2vf = 133.9 m/s = 130 m/s

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Sample Problems

1. A ball rolling down a hill at 4.0 m/s accelerates at 2.0 m/s2 What is its velocity 5.0s later

Given:v1 = 4.0 m/sa = 2.0 m/s2t = 5.0s

We can use equation 1Equation 1: v2 = v1 + at

v2 = 4.0m/s + 2.0 m/s2*5.0s

v2 = 4.0 m/s + 10.0 m/s = 14.0 m/s

The ball reaches a velocity of 14 m/s in 5.0 s.

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2. A car travelling at 10 m/s (2 Sig figs) accelerates at 4.0 m/s2 for 8.0s. What is its displacement during this interval?

Given v1 = 10 m/sa = 4.0 m/s2t = 8.0s find Δd

We can use equation # 3


Δx =vit + 1/2 at2

=10 m/s* 8.0s + 1/2 * 4.0m/s2 * 8.0s2

= 80m + 128m = 208 m = 2.1 x 102m

The cars displacement for 8.0s is 2.0 x 102 m

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3.A car accelerating at 5.0 m/s2 has a displacement of 114m in 6.0s. What was its velocity at the beginning of the interval?

given:

a = 5.0 m/s2t = 6.0sΔd = 114

find v1

We can use equation 3


Δd = xf - xi

Δd = vit + 1/2at2

114m = vi(6.0s) + 1/2 (5.0 m/s2)(6.0s)2

114m = 6.0s(vi) + 90 mcombine like terms

114m - 90m = 6.0s (vi)24m = 6.0s (vi)

vi = 4.0 m/s

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4. A ball rolls at an initial velocity of 4.0 m/s up a hill. five seconds later it is rolling down the hill at 6.0 m/s2.

Find the following:

a) accelerationb) displacement at 5.0 s.

a) assuming up the hill is positive and down the hill is negative (in terms of displacement and therefor speed) given: v1 = 4.0 m/sv2 = 6.0 m/st = 5.0s

find aa = (v2 v1)/Δt = (6.0 m/s 4.0 m/s)/5.0s = 10m/s/5s = 2m/s2

b) Use a to find displacementEquation #2 or # 4


2 + 2aΔx




2 + 2aΔx

Equation 1: v2 = v1 + at


Using equation 2Δx =( 1/2)(v1 +v2)*t0.5* (-6.0m/s +4.0m/s)(5.0s)= 0.5(-2.0m/s)(5.0s)=-5m

The ball is 5.0 m down the hill from its starting pint after 5.0s

Using equation 4(-6.0m/s)2 = (4.0m/s)2 + 2(-2m/s2)(Δx)36 = 16 + -4x20 = -4xx = -5m

uniform rectilinear motion uniform accelerated...

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