Kinematics Acceleration Objectives Define acceleration and depict acceleration
Kinematics Acceleration
Objectives • Define acceleration and depict acceleration using • written sentences • a motion diagram • a data table • graphs of position versus time, velocity versus time, and acceleration versus time. • Describe the difference between instantaneous, average, and constant velocity. • Determine the magnitude and direction of acceleration from a positiontime graph, a velocity-time graph, and an acceleration-time graph. • Describe how an object can have constant speed and be accelerating. • Describe how and object’s acceleration can be decreasing while the object is speeding up.
Acceleration is the rate of change in velocity a= v t Acceleration is NOT the rate of velocity. It is the rate of CHANGE in velocity. Since velocity is the rate of change in position (displacement), this means that acceleration is the rate of a rate. Acceleration is not how fast an object is moving (that’s velocity). Acceleration is a measure of how quickly is an object’s velocity changing Units: Velocity m/s Time s Acceleration: m/s/s = m/s 2
Types of Speed and Velocity • Constant speed Speed that is not changing. Remember speed is scalar and has magnitude only. Constant speed makes no mention of direction. • Constant velocity Velocity that is not changing, v = 0. Velocity is a vector consisting of both magnitude and direction. Constant velocity must have both constant magnitude (constant speed) and constant direction. This implies that acceleration is zero: a = 0 • Instantaneous speed and velocity Speed and velocity at a specific instant in time. In acceleration problems velocity may be changing, but at a specific instant of time there will be a specific instantaneous value for speed and velocity. • Average speed: The total distance traveled divided by the total time. • Average velocity: Displacement divided by the total time, vavg = x / t This seems the same as the constant velocity equation. What’s the difference?
For Objects at Constant Velocity Must they have constant velocity? Yes: Constant velocity means constant magnitude and constant direction. For linear motion speed has the same magnitude as velocity. If velocity is constant, then speed is always constant. Speed is a scalar. It is found by dividing the distance scalar by the elapsed time. Velocity is a vector. It is found by dividing the displacement vector by the elapsed time. While speed and velocity are completely different for non-linear motion, they have the same magnitudes in linear motion. Can they be accelerating? No: Acceleration requires a change in velocity during a time interval, and constant velocity has constant magnitude and direction. If there is no change in velocity, then acceleration is always zero. Zero values in word problems are often indicated by the wording in the problem. The words “constant velocity” give you an important zero quantity, a = 0.
For Objects at Constant Speed Must they have constant velocity? No: An object at constant speed might have constant velocity, as long as it is traveling in a straight line. However, an object moving with constant magnitude along a curved path will have constant speed, but not constant velocity (Constant velocity needs both magnitude and direction to be constant). Look at the wording in this problem: It says “Must they have constant velocity? ” The word “must” is very restrictive. To answer, yes, there can be no possible exceptions. A turning object has constant speed, but not constant velocity. This is an exception to the question, and the answer has to be “no”. Can they be accelerating? Yes: An object at constant speed might be turning. A turning object is experiencing a change in direction, which changes velocity (a vector quantity with both magnitude and direction). Acceleration is the rate of change in velocity. If velocity is changing during a time interval, then the object is accelerating. This time the question says “Can they be accelerating? ” The word “can” is not as restrictive as the word “must. ” In this case we are looking for any exceptions. If we find even one exception, then the answer is “yes”.
Galileo’s Incline: A classic example of acceleration When a ball is rolled down an incline with an angle of 12. 84 o the acceleration of the ball will be 2 m/s 2. (How this is determined will be explained later. ) A table quantifying this motion for four seconds would appear as follows: Time (s) Position (m) Velocity (m/s) Acceleration (m/s 2) 0 0 0 2 1 1 2 2 2 4 4 2 3 9 6 2 4 16 8 2 • Constant acceleration: Constant magnitude and constant direction (2, 2, 2, …). • If a = 2 m/s 2 , then velocity is changing by 2 m/s every 1 s (0, 2, 4, 6, …). • Displacement is proportional to the square of time x = ½ a t 2 (0, 1, 4, 9, …). Note: displacement and position have the same value when objects start at the origin (x 0 = 0). x = x x 0 x = x
Galileo’s Incline A motion diagram of the ball, depicting the balls position at every second, would appear as follows. When objects experience constant velocity they move the same distance every second and spacing between the dots in a motion diagram remains constant. However, when object’s accelerate velocity is changing. A change in velocity will then change subsequent distance moved by the object. When objects accelerate the spacing between dots in a motion diagram varies. Here the velocity of the object is increasing. If the velocity increases each second, then the distance moved by the object is greater in each subsequent second. When an object slows down the space between dots in a motion diagram decreases.
Galileo’s Incline 1 m 3 m 5 m 7 m When Galileo measured the space between the bells on his incline (see the explanation of his experiment in the prior lab) he noticed that the spacing was proportional to the odd numbers. He called this the Rule of Odd Numbers. When an incline of 12. 84 o is used, then the interval distances work out to be the odd numbers in meters every second. While an interesting pattern it is not as important as the displacements measured from the starting location.
Galileo’s Incline 1 m 3 m 5 m 4 m 7 m 9 m 16 m Displacement is directly proportional to the square of time: x t 2 The equation relating displacement, acceleration, and time is: x = ½ a t 2 For an incline of 12. 84 o acceleration is 2 m/s 2. When substituted into the above equation it simplifies to x = t 2. As a result, the equality now directly matches the proportionality. This makes seeing the numerical trends and proportionality easier.
Galileo’s Incline 1 m 3 m 5 m 0 5 7 m 10 9 m 15 t (s) x (m) Δx (m) v (m/s) t 0 = 0 x 0 = 0 0 v 0 = 0 1 1 1 2 2 4 4 4 3 9 9 6 4 16 16 8 5 25 25 10 6 36 36 12 11 m 20 25 30 This motion can be viewed several ways. • As a motion diagram • In a table • In a positon versus time graph • In a velocity versus time graph 12 10 40 8 30 v (m/s) x (m) 20 6 4 10 2 0 0 0 2 4 t (s) 6 8 10 35 0 2 4 t (s) 6 8 10 40
Galileo’s Incline 1 m 3 m 5 m 0 5 7 m 10 9 m 15 t (s) x (m) Δx (m) v (m/s) t 0 = 0 x 0 = 0 0 v 0 = 0 1 1 1 2 2 4 4 4 3 9 9 6 4 16 16 8 5 25 25 10 6 36 36 12 11 m 20 25 30 Now, however we can add a column to the table for acceleration, and we can add an acceleration versus time graph. 10 40 8 30 v (m/s) 6 4 10 2 0 0 0 2 4 t (s) 6 8 10 40 The slope of position versus time is still velocity, and the area of velocity versus time is still displacement. 12 x (m) 20 35 0 2 4 t (s) 6 8 10
Galileo’s Incline 1 m 3 m 5 m 0 5 7 m 10 9 m 15 11 m 20 t (s) x (m) Δx (m) v (m/s) a (m/s 2) t 0 = 0 x 0 = 0 0 v 0 = 0 a 0 = 2 1 1 1 2 2 2 4 4 4 2 3 9 9 6 2 4 16 16 8 2 5 25 25 10 2 6 36 36 12 2 25 30 35 40 Since velocity is changing the slope of position versus time is also changing. Instantaneous velocity, at a specific time, is the slope of a line tangent to the function at that instant. (Differential calculus). 12 10 40 8 30 8 6 6 x (m) 20 v (m/s) 4 10 2 2 0 0 2 4 t (s) 6 8 10 a (m/s 2) 0 2 4 t (s) 6 8 10 4 0 2 4 t (s) 6 8 10
Galileo’s Incline 1 m 3 m 5 m 0 5 7 m 10 9 m 15 11 m 20 t (s) x (m) Δx (m) v (m/s) a (m/s 2) t 0 = 0 x 0 = 0 0 v 0 = 0 a 0 = 2 1 1 1 2 2 2 4 4 4 2 3 9 9 6 2 4 16 16 8 2 5 25 25 10 2 6 36 36 12 2 25 30 35 40 Since velocity is changing the slope of position versus time is also changing. Instantaneous velocity, at a specific time, is the slope of a line tangent to the function at that instant. (Differential calculus). 12 10 40 8 30 8 6 6 x (m) 20 v (m/s) 4 10 2 2 0 0 2 4 t (s) 6 8 10 a (m/s 2) 0 2 4 t (s) 6 8 10 4 0 2 4 t (s) 6 8 10
Galileo’s Incline 1 m 3 m 5 m 0 5 7 m 10 9 m 15 11 m 20 t (s) x (m) Δx (m) v (m/s) a (m/s 2) t 0 = 0 x 0 = 0 0 v 0 = 0 a 0 = 2 1 1 1 2 2 2 4 4 4 2 3 9 9 6 2 4 16 16 8 2 5 25 25 10 2 6 36 36 12 2 25 30 35 40 Now examine the slope of velocity vs time. The rise is a change in velocity and the run is elapsed time. The slope of velocity vs time is acceleration. 12 10 40 8 30 8 6 6 x (m) 20 v (m/s) 4 10 2 2 0 0 2 4 t (s) 6 8 10 a (m/s 2) 0 2 4 t (s) 6 8 10 4 0 2 4 t (s) 6 8 10
Galileo’s Incline 1 m 3 m 5 m 0 5 7 m 10 9 m 15 11 m 20 t (s) x (m) Δx (m) v (m/s) a (m/s 2) t 0 = 0 x 0 = 0 0 v 0 = 0 a 0 = 2 1 1 1 2 2 2 4 4 4 2 3 9 9 6 2 4 16 16 8 2 5 25 25 10 2 6 36 36 12 2 25 30 35 40 Examine the area of acceleration vs time. The height is acceleration and the base is elapsed time. The area of acceleration versus time is change in velocity. 12 10 40 8 30 8 6 6 x (m) 20 v (m/s) 4 10 2 2 0 0 2 4 t (s) 6 8 10 a (m/s 2) 0 2 4 t (s) 6 8 10 4 0 2 4 t (s) 6 8 10
Galileo’s Incline 1 m 3 m 5 m 0 5 7 m 10 9 m 15 11 m 20 t (s) x (m) Δx (m) v (m/s) a (m/s 2) t 0 = 0 x 0 = 0 0 v 0 = 0 a 0 = 2 1 1 1 2 2 2 4 4 4 2 3 9 9 6 2 4 16 16 8 2 5 25 25 10 2 6 36 36 12 2 25 30 35 40 How can v be used to find the instantaneous velocity, v , at a specific time? 12 10 40 8 30 8 6 6 x (m) 20 v (m/s) 4 10 2 2 0 0 2 4 t (s) 6 8 10 a (m/s 2) 0 2 4 t (s) 6 8 10 4 0 2 4 t (s) 6 8 10
Galileo’s Incline 1 m 3 m 5 m 0 5 7 m 10 9 m 15 11 m 20 25 30 t (s) x (m) Δx (m) v (m/s) a (m/s 2) v = Area a-t + v 0 t 0 = 0 x 0 = 0 0 v 0 = 0 a 0 = 2 0 = 0 +0 1 1 1 2 2 2 = 2 +0 2 4 4 4 2 4 = 4 +0 3 9 9 6 2 6 = 6 +0 4 16 16 8 2 8 = 8 +0 5 25 25 10 2 10 = 10 +0 6 36 36 12 2 12 = 12 +0 12 35 40 Area is measured from the origin to time, t. The initial velocity must be given. In this case, v 0 = 5 m/s. Area accumulates over time. 10 40 8 30 8 6 6 x (m) 20 v (m/s) 4 10 2 2 0 0 2 4 t (s) 6 8 10 a (m/s 2) 0 2 4 t (s) 6 8 10 4 0 2 4 t (s) 6 8 10
Galileo’s Incline 1 m 3 m 5 m 0 5 7 m 10 9 m 15 11 m 20 25 30 t (s) x (m) Δx (m) v (m/s) a (m/s 2) x = Area v-t + x 0 t 0 = 0 x 0 = 0 0 v 0 = 0 a 0 = 2 0 = 0 +0 1 1 1 2 2 1 = 1 +0 2 4 4 4 2 4 = 4 +0 3 9 9 6 2 9 = 9 +0 4 16 16 8 2 16 = 16 +0 5 25 25 10 2 25 = 25 +0 6 36 36 12 2 36 = 36 +0 35 40 In a similar manner the area of velocity versus time can be used to find displacement and then position. The initial position must be given, and is x 0 = 0 m. 12 10 40 8 30 8 6 6 x (m) 20 v (m/s) 4 10 2 2 0 0 2 4 t (s) 6 8 10 a (m/s 2) 0 2 4 t (s) 6 8 10 4 0 2 4 t (s) 6 8 10
Overview of how the graphs relate to each other x a v t t t
Acceleration is complicated both mathematically and graphically Acceleration has magnitude and direction. Magnitude is tied to change (Δ) and can be positive or negative. Vector direction can also be positive or negative. When acceleration is changed into a scalar quantity for calculation purposes there are four possibilities resulting from the combination of these signs. Motion to right (+) , velocity increasing (+) Motion to right (+) , velocity decreasing (−) Motion to left (−) , velocity increasing (+) Motion to left (−) , velocity decreasing (−)
How does the sign on acceleration affect graphs? Speeding up Right ++a x Slowing Right −+a Speeding up Left +−a x t v t a t Slowing Left −−a t t
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