Presentation on theme: "Acceleration. Velocity (v) - rate of position change. Constant v – rate stays the same, equal distance for equal t interval. Acceleration (a)- rate of."— Presentation transcript:
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:
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
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?
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 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#