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

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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 =  va = v f – v i 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?

Units Since a =  v /t Units become:  m  v s = m or d/t 2.  t ss 2 What does it mean to have a constant or uniform acceleration of = 10 m/s 2.

Acceleration of Gravity (g) near the surface of the Earth. ~ 10 m/s 2. Drop a ball from rest (v i = 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 positive since  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 = v f –v i t a  v +20 m/s - 0 m/s= +10 m/s 2  t2 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 v f is 0, and v i is +20m/s, so. a  v 0 m/s - (+20 m/s)= -10 m/s 2.  t2s 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 v f is neg. Hold on - its not so simple!

A quick calculation shows this new accl to be negative. Oy! Here are the rules:

Velocity motiona posspeeding+ posslowing- negspeeding- negslowing+ Mental Trick!

4. A shuttle bus slows to a stop with an acceleration of -1.8 m/s 2. How long does it take to slow from 9.0 m/s to rest? List the variables. a = -1.8 m/s 2. v i = 9 m/s v f = 0 (stop) t = ?

Find an equation with everything on the list. a = v f – v i /t Rearrange to solve for the unknown. t = v f – v i /a Plug in with units. = 9m/s – 0 =5 s. -1.8 m/s 2 a = -1.8 m/s 2. v i = 9 m/s v f = 0 (stop) t = ?

7200 m/s. 5. A plane starts from rest and accelerates for 6 minutes at 20 m/s 2 before traveling at a constant velocity. What was its final velocity?

Another useful acceleration equation. d = v i t + ½ at 2. d= displacement (m) v i = initial (starting velocity) m/s. a = acceleration m/s 2. t = time over which accl takes place - s.

6. A car starts from rest and accelerates at 5 m/s 2 for 8 seconds. How far did it go in that time? List! Equation! Solve. 160 m

Acceleration Hwk. Hwk Intro to Acceleration Rd 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 v f = v i + at d = v i t + ½ at 2. v f 2 = v i 2 + 2 ad.

1. A car starts from rest and accelerates at 6 m/s 2 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. v i = 23 m/s v f = 0 d = 23 m a = ? v f 2 = v i 2 + 2ad - v i 2 = a 2d -(23 m/s) 2 = -22 m/s 2. 2(12m)

3.A bicycle is traveling at 5 m/s. It accelerates at 3 m/s 2 for 3 meters. What is its final velocity?

4. A truck skidded to a stop with an acceleration of -3 m/s 2. 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 Graphs Speed 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 Book Hwk Rev Book. Rd 55 – 58. Do pg 56 #7-13 AND Pg 58 #14 – 18.

Displacement on V-T graphs

Displacement = Area Under Curve v = 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 –v f – v i.

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/s 2 – very close to 10 m/s 2.

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 Weightlessness Objects 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/s 2. 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/s 2. 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?

v i = +25 m/s a = g = -9.81 m/s 2. t = 5 s. d = ? d= v i t + ½ at 2. (25m/s)(5s) + 1/2(-9.81 m/s 2 )(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 = v i t + ½ at 2. -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/s 2. v f = ? Hmmm v i = 0. v f 2 = v i 2 + 2ad v f 2 = 2(-9.81m/s 2 )(7 m) v f = -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)dv (m/s)a (m/s 2 ) 0025-9.81 12015.2-9.81 2305.4-9.81 331-4.43-9.81 52.4-24-9.91

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#

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