Acceleration

Velocity (v) - rate of position change. Constant v – rate stays the same, equal distance for equal t interval.

Acceleration (a)- rate of velocity change.

Constant/Uniform accl. – change v for equal time interval.

To calculate average/ uniform / constant accl

a = v a = vf – vi

t t

Velocity vs. Acceleration

Acceleration

• Any object not traveling at constant velocity is accelerating.

• Changing speed – either speeding up or slowing down.

• Changing direction.

We will look at

Uniform / Constant Acceleration

Changing speed or direction at constant rate.

What are units?

Since a = v /tUnitsUnits become:

mv s = m or d/t2.t s s2

What does it mean to have a constant or uniform acceleration of = 10 m/s2.

Acceleration of Gravity (g) near the surface of the Earth.

• ~ 10 m/s2.

• Drop a ball from rest (vi = 0)

• After 1 sec, it’s velocity ~ 10 m/s

• After 2 sec, it’s velocity ~

a. If I drop a ball from rest near Earth’s surface, approximately how fast is it moving after falling 5 seconds?

• 50 m/s

Acceleration is a vector.Magnitude & direction.

When v is positive accl is positivesince v

t

What does the sign of accl mean?

Sign of acceleration

1. Consider a car starts from rest, heads east, and attains a velocity of +20 m/s in 2 s. Calculate a:

Sign tells change in v.

a = vf –vi

t av +20 m/s - 0 m/s = +10 m/s2

t 2 s

Acceleration is positive.

The car is headed in a positive direction and speeding up.

2. Now consider the same car slowing to a stop from +20 m/s in 2s. Calculate a.

Now vf is 0, and vi is +20m/s, so.

av 0 m/s - (+20 m/s) = -10 m/s2.t 2s

Accl is neg. the car is slowing down.

3. Consider the same car starts from rest, heads west, and reaches 20 m/s in 2 s. Calculate a now. Since the car is heading west, the vf is neg.

Hold on - its not so simple!

A quick calculation shows this new accl to be negative. Oy!

Here are the rules:

Velocity motion a

pos speeding +pos slowing -neg speeding -neg slowing +

Mental Trick!

4. A shuttle bus slows to a stop with an acceleration of -1.8 m/s2. How long does it take to slow from 9.0 m/s to rest?

• List the variables.

• a = -1.8 m/s2.

• vi = 9 m/s

• vf = 0 (stop)

• t = ?

• Find an equation with everything on the list.

• a = vf – vi/t

• Rearrange to solve for the unknown.

• t = vf – vi/a

• Plug in with units.• = 9m/s – 0 = 5 s.

-1.8 m/s2

• a = -1.8 m/s2.

• vi = 9 m/s

• vf = 0 (stop)

• t = ?

• 7200 m/s.

5. A plane starts from rest and accelerates for 6 minutes at 20 m/s2 before traveling at a constant velocity. What was its final velocity?

Another useful acceleration equation.

• d = vit + ½ at2.

• d= displacement (m)

• vi = initial (starting velocity) m/s.

• a = acceleration m/s2.

• t = time over which accl takes place - s.

6. A car starts from rest and accelerates at 5 m/s2 for 8 seconds. How far did it go in that time?

• List!

• Equation!

• Solve.

• 160 m

Acceleration Hwk.

• Hwk Intro to AccelerationRd 48 – 49 Do pg 49 #1 - 5.

• 1. Take your seat.

• 2. Take out your physics supplies.

• 3. Will 6 volunteers come to the front please

Do Now:

• Ruler Drop.

More Acceleration Equations

• a = v/t

• vf = vi + at

• d = vit + ½ at2.

• vf2 = vi

2 + 2 ad.

1. A car starts from rest and accelerates at 6 m/s2 for 5 seconds. How far did it travel?

2. A car slows to a stop from 23 m/s by applying the brakes over 12 meters. Calculate acceleration.

• vi = 23 m/s

• vf = 0

• d = 23 m• a = ?

• vf2 = vi

2 + 2ad

• - vi2 = a

• 2d• -(23 m/s)2 = -22 m/s2.

2(12m)

3.A bicycle is traveling at 5 m/s. It accelerates at 3 m/s2 for 3 meters. What is its final velocity?

4. A truck skidded to a stop with an acceleration of -3 m/s2. If its initial velocity was 11 m/s, how far was it skidding before it came to a stop?

• Hwk Wksht Mixed Accl Equations sheet.

Velocity Time GraphsSpeed Time Graphs

Constant Velocity/Speed

Constant / Uniform Acceleration. On velocity time graph accl. is slope of straight line.

What’s going on here?

Sign of velocity is direction.

Sketch Graphs

V-t sketch graphs Rev BookHwk Rev Book. Rd 55 – 58. Do

pg 56 #7-13 AND Pg 58 #14 – 18.

Displacement on V-T graphs

Displacement = Area Under Curvev = d/t then, vt = d.

Area of non-constant velocity. For constant accl, d = area of a triangle: ½ bh.If a car achieved a v=40 m/s in 10 s, then:½(10s)(40m/s) = 200 m.

40 m/s

10 s

To find displacement, calc area of triangle + rectangle.

How can you tell when object is back to starting point?

• Positive displacement = negative displacement.

Do Now: Given the v – t graph below, sketch the acceleration – t graph for the same motion.

Acceleration – time Graphs

• What is the physical behavior of the object?• Slowing down pos direction, constant vel neg accel.

• d-t: – slope = velocity– area ≠ .

• v-t: – slope = accl– area = displ

• a-t: – slope ≠ .– area = vel

– vf – vi.

Hwk Rev Book pg 76 #32-36, 45-47, 49 – 55, 59-60. Begin in class.

Together: text pg 65 #1-5,pg 72 # 34, 35, 48, 49.

Test Thursday: Acceleration, Motion Graphs, Free-fall. Text 2-2, 2-3. RB Chap 3.

Objects Falling Under Gravity

Freefall

Gravity accelerates uniformly masses as they fall and rise.

Earth’s acceleration rate is 9.81 m/s2 – very close to 10 m/s2.

Falling objects accelerate at the same rate in absence of air resistance

But with air

resistance

Fortunately there is a “terminal fall velocity.”

After a while, the diver falls with constant velocity due to air resistance.

Unfortunately terminal fall velocity is too large to live through the drop.

Apparent WeightlessnessObjects in Free-fall Feel Weightless

What is the graph of a ball dropped?

What do the d-t, v-t, and a – t, graphs of a ball thrown into the air look like

if it is caught at the same height?

A ball is thrown upward from the ground level returns to same height.

d = ball’s height above the ground

velocity is + when the ball is moving upward

Why is acceleration negative?

a is -9.81, the ball is accelerating at constant 9.81 m/s2.

Is there ever deceleration?

Free-fall Assumptions

Trip only in the air. Trip ends before ball caught.

-Symmetrical Trip time up = time down

-Top of arc:

v = 0,

a = ??

-On Earth g = -9.81 m/s2. Other planets g is different.

Solving:Use accl equations replace a with -g.

• List given quantities & unknown quantity.

• Choose accl equation that includes known & 1 unknown quantity.

• Be consistent with units & signs.

• Check that the answer seems reasonable

• Remain calm

Practice Problem.• 1. A ball is tossed

upward into the air from the edge of a cliff with a velocity of 25 m/s. It stays airborne for 5 seconds. What is its total displacement?

vi = +25 m/sa = g = -9.81 m/s2. t = 5 s.d = ?

• d= vit + ½ at2. • (25m/s)(5s) + 1/2(-9.81 m/s2)(5 s)2. • 125 m - 122 . 6 = +2.4m.

• It is 2.4m above the start point.

2. If the air time from the previous problem is increased to 5.2 seconds, what will be the displacement?

• d = vit + ½ at2.

• -2.6 m

• It will be below the start point.

Ex 3. A 10-kg rock is dropped from a 7- m cliff. What is its velocity just before hitting the ground?

• d = 7m• a = -9.81 m/s2.

• vf = ?

• Hmmm

• vi = 0.

• vf 2 = vi2 + 2ad

• vf 2 = 2(-9.81m/s2)(7 m)

• vf = -11.7 m/s (down)

 4. A ball is thrown straight up into the air with a velocity of 25 m/s. Create a table showing the balls position, velocity, and acceleration for each second for the first 5 seconds of its motion.

•  T(s) d v (m/s) a (m/s2)

• 0 0 25 -9.81

• 1 20 15.2 -9.81

• 2 30 5.4 -9.81

• 3 31 -4.43 -9.81

• 5 2.4 -24 -9.91

Read Text pg 60-64 Do prb’s pg 64# 1-5 show work.

Mech Universe: The Law of Falling Bodies:http://www.learner.org/resources/series42.html?pop=yes&pid=549#