Motion InClass Assignment 1 Break into groups of

Motion

In-Class Assignment 1) Break into groups of 2 to 3 people. 2) Using your domino, execute a simple motion. 3) On a sheet of paper, write as accurately as possible a description of the motion. Write your group members’ names on the paper and turn it in when you are complete. This should take no longer than 10 minutes.

Would the Motions We Observe Everyday Appear Differently if Observed from a Different Vantage Point?

Motion is Relative …To the position of the observer. I collided with a stationary truck coming the other way. A pedestrian hit me and went under my car. The guy was all over the road. I had to swerve a number of times before I hit him. The telephone pole was approaching fast. I was attempting to swerve out of its path when it struck my front end.

Just For Fun… "In an attempt to kill a fly, I drove into a telephone pole. " "Coming home I drove into the wrong house and collided with a tree I don't have. " "I thought my window was down, but I found it was up when I put my head through it. " "To avoid hitting the bumper of the car in front I struck a pedestrian. " "I was sure the old fellow would never make it to the other side of the road when I struck him. " "The pedestrian had no idea which way to run as I ran over him. " "The pedestrian ran for the pavement, but I got him. "

What terms are necessary to accurately describe the motion of ANY object?

Position 0 1 2 3 4 5 Position (x): location, can be specified in one, two or three dimensions, described in terms of a distance from an origin (reference point), given in units of length.

Displacement 0 1 2 3 4 5 displacement: a change in position. Often denoted by s or d or x, displacement has units of length, and is a vector quantity. xi xf

Distance vs. Displacement A B

Displacement is a Vector Quantity Vector quantities require both a magnitude AND a direction to fully describe them. 25 meters North Magnitude Direction Distance is a scalar quantity.

Displacement Can Be Negative 0 5 10 15 Which just indicates displacement in the negative direction.

Position vs. Time Graph What’s the total displacement? Position (m) 10 (a) 10 meters (b) 0 meters (c) 5 meters (d) 15 meters 5 0 10 5 Time (s) 15

Velocity Average velocity (vavg) – the total displacement divided by the time interval during which the displacement occurred. Instantaneous velocity – the velocity at a particular instant. Speed is a scalar quantity. Velocity is a vector quantity. Speed is the magnitude of velocity…

Average Velocity Vavg = 5 miles/0. 2 hours = 25 miles/hour

Position vs. Time Graph Position (m) 10 5 x 0 t 10 5 Time (s) 15

The Slope of the Position vs. Time Graph is the Velocity Positive slopes indicate positive velocities, negative slopes, negative velocities. x (m) t (s)

The Slope of the Position vs. Time Graph is the Velocity Constant slopes indicate constant velocities. x (m) t t t x x x t (s)

The Slope of the Position vs. Time Graph is the Velocity Increasing slopes indicate increasing velocities, decreasing slopes indicate decreasing velocities. x (m) t (s)

Concept Check… Get Your Cards…

The graph represents the motion of two cars. Which statement is correct? (a) The blue car is slower than the red car. (b) The cars have the same speed at t = 12 sec. (c) The red car is speeding up. (d) All roads lead to Rome. x(m) 20 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 t (s)

(a) The blue car is slower than the red car. The slope of the red car is steeper/greater than that of the blue car, therefore, the velocity is likewise greater. x(m) 20 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 t (s)

This graph shows (a) speeding up in the positive direction. (b) slowing down in the negative direction. (c) speeding up in the negative direction. (d) slowing down in the positive direction. x(m) 20 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 t (s)

(c) speeding up in the negative direction. The slope begins at nearly zero, and becomes increasingly more negative. x(m) 20 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 t (s)

This graph shows (a) speeding up in the positive direction. (b) slowing down in the negative direction. (c) speeding up in the negative direction. (d) slowing down in the positive direction. x(m) 20 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 t (s)

(b) slowing down in the negative direction. The graph begins with a fairly steep negative slope which becomes less and less steep, nearly zero, at the end of 15 seconds. x(m) 20 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 t (s)

Average Velocity from a Graph Position (m) 10 5 0 x t 10 5 Time (s) 15 Which really is the slope of the line between the two positions.

Instantaneous Velocity Position (m) The instantaneous velocity for any time t may be pulled from a position vs. time graph as the slope of the graph at that time. 10 t x 5 x 0 t 5 10 15 Time (s)

The Velocity vs. Time Graph… Your new best friend…

The Velocity vs. Time Graph Velocity (m/s) 10 5 0 10 5 Time (s) 15

Acceleration Average acceleration (aavg) is the change in velocity divided by the time interval during which the change occurred. Instantaneous acceleration is the acceleration at a particular instant. Acceleration is a vector quantity.

Units for Acceleration The units for acceleration have the dimensions of length over time squared: Very often, we’ll report accelerations in meters per second squared (m/s 2), but any other combination of units that have the same dimensions are acceptable.

What Does an Acceleration of 10 m/s 2 Mean? It means that the velocity of an object is changing at a rate of 10 m/s every second.

The Velocity vs. Time Graph Velocity (m/s) 10 5 v 0 t 10 5 Time (s) 15

The Slope of the Velocity vs. Time Graph is the Acceleration v (m/s) Positive slopes indicate positive accelerations, negative slopes, negative accelerations. 0 t (s)

The Slope of the Velocity vs. Time Graph is the Acceleration v (m/s) Constant slopes indicate constant acceleration. t t v v t (s)

The Slope of the Velocity vs. Time Graph is the Acceleration Increasing slopes indicate increasing accelerations, decreasing slopes indicate decreasing velocities. v (m) t (s)

Something to Consider… Velocity (m/s) Let’s say that you are traveling (graph below) southbound with a constant velocity of 10 m/s. If you travel for 4 seconds, how far have you traveled? How far have you traveled in 6 seconds? 10 40 m = 10 m/s x 4 s 0 2 4 6 Time (s)

Something Else to Consider… Velocity (m/s) Let’s say that you are start from rest and travel (graph below) southbound with a constant acceleration of 1 m/s 2. If you travel for 4 seconds, how far have you traveled? How far have you traveled in 6 seconds? 10 vavg= (vf – vi)/2= 2 m/s x 4 s = 8 m = ½ bh = ½ (4 s)(4 m/s) 0 4 2 Time (s) 6

What a Revelation!!!!! The area between the graph and the horizontal axis is the displacement that occurred during that time interval.

A Second Look Velocity (m/s) 10 A 1 = ½ (5 m/s)(5 s) = 12. 5 m A 2 = =(5 m/s)(5 s) = 25 m A 3 = ½ (5 m/s)(5 s) = 12. 5 m 5 A 2 A 1 0 A 3 10 5 Time (s) 15 x(m) t(s) 0 0 12. 5 5 37. 5 10 50 15

Position (m) You Can Draw an X vs. T Graph from a V vs. T Graph (50, 15) 50 (10, 37. 5) 25 (5, 12. 5) 0 5 10 x(m) t(s) 0 0 12. 5 5 37. 5 10 50 15 15 Time (s)

The graph represents the motion of two cars. Which statement is correct? v(m/s) (a) The blue car is always slower than the red car. (b) The cars have the same speed at t = 12 sec. (c) The cars have the same position at r = 12 sec. (d) The blue car has greater acceleration. 20 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 t (s)

A rock is dropped from a hovering helicopter. If it’s initial velocity is 0 m/s, and the acceleration due to gravity is 10 m/s 2, what is the velocity of the rock after it has fallen for 5 seconds? (a) 0 m/s (b) 5 m/s (c)50 m/s (d)20 m/s v=?

(c) 50 m/s Every second the rock falls, its velocity increases by 10 m/s. t=0 t=1 s v=0 v = 10 m/s t=2 s v = 20 m/s t=3 s v = 30 m/s t=4 s v = 40 m/s t=5 s v = 50 m/s

(b) The cars have the same speed at t = 12 sec. v(m/s) At t = 12 sec, the cars have the same speed. 20 18 16 14 12 10 8 6 4 2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 t (s)

From 0 – 6 sec, the graph shows v(m/s) (a) speeding up in the positive direction. (b) slowing down in the negative direction. (c) speeding up in the negative direction. (d) slowing down in the positive direction. 12 10 8 6 4 2 0 -2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -4 -6 t (s)

(d) slowing down in the positive direction. v(m/s) The object’s motion begins with a velocity of 10 m/s and decreases at a constant rate to 0 m/s at t = 6 sec. 12 10 8 6 4 2 0 -2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -4 -6 t (s)

From 0 – 6 sec, v(m/s) (a) The velocity and acceleration are positive. (b) The velocity and acceleration are negative. (c) The velocity is positive, and acceleration negative. (d) The velocity is negative, and acceleration positive. 12 10 8 6 4 2 0 -2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -4 -6 t (s)

(c) The velocity is positive, and acceleration negative. v(m/s) The slope (acceleration) is negative, and the velocity is positive. 12 10 8 6 4 2 0 -2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -4 -6 t (s)

Another Case of Acceleration Since acceleration is defined as a change in velocity over time, and velocity is a vector quantity, a change in direction constitutes acceleration also. This is called centripetal acceleration, and will be discussed later

Observe the animation of the three cars below. Which car or cars (red, green, and/or blue) are undergoing an acceleration? Study each car individually in order to determine the answer.