MOTION ALONG A STRAIGHT LINE Back and forth

MOTION ALONG A STRAIGHT LINE Back and forth, or up and down; Or north to south along the ground. When things move, or just fall-free, Take delta ‘x’ by delta ‘t’ To get average VELOCITY

RELATIVE MOTION Direction of motion An observer on shore sees the ball falling sideways An observer on the ship sees the ball fall straight down

SIMULTANEITY (RELATIVE TIME) U

FREE FALL – A special case n Galileo demonstrated that, in the absence of air resistance, objects will fall at the same rate near the surface of the earth n Assuming Galileo is correct, sketch a graph of distance, velocity, and acceleration versus time for an object dropped from the top of the Leaning Tower of Pisa

Freefall

SECOND OPINION

GRAPHS OF MOTION The easiest way to start, is to identify a frame of reference and any constant quantities. n For this event, where is a good place to start and what remains constant as the object falls? n How does this information help produce the other graphs? n

Motion Graphs y=+ y=0 Acceleration = constant Make axes and sketch the graphs y=-

PROBLEM ALERT! n How is it possible to determine the velocity of an object for one instant in time?

Instantaneous velocity as a limit n n Suppose the position as a function of time is given by: x(t)=5 t 2 Sketch the graph of position vs. time for the object from zero to six seconds n n n Determine the velocity of the object at four seconds The instantaneous velocity may be determined by making smaller and smaller tangent lines Therefore, the expression for the instantaneous velocity becomes:

GRAPHING MOTION CHALLENGE n Suppose a graph showing velocity versus time is not a straight line, how could it be used to produce graphs of displacement and acceleration?

MISCONCEPTION ALERT! n The ‘BIG FIVE’ kinematics equations (see table 2 -1) are only valid for situations involving constant acceleration!!

ADVENTURES IN CALCULUS n n Distance as an integral To determine the distance traveled using a velocity vs. time graph, calculate the area under the curve n n n Acceleration as a derivative To calculate the acceleration, determine the derivative Remember, a derivative is simply the slope of a graph!

Distance as an integral Suppose the velocity of a particle in straight line motion is given by the equation: V(t) = 5 + 6 t n Make a graph of v vs. t for 0 – 5 seconds. n From the graph, calculate the area under the curve for t=4 s n

Distance as an integral

Distance as an integral n n n Since an integral is the same as calculating the area under a curve, the distance traveled is equal to the area calculated. An integral is also an anti – derivative. In other words – it is a function whose derivative gives back the original function. For this problem, what function’s derivative will yield 5+6 t? Try to work backwards using the power rule

Distance as an integral n The power rule for integration is: By applying this to v(t) = 5+6 t, the integral is: X(t) = 5 t +3 t 2 With the insertion of limits, the result may be calculated:

Distance as an integral [5(4)+3(4)2] – [5(0)+3(0)2] = 68 Compare this to the area calculation!

Acceleration as a derivative Suppose the velocity of a particle in straight line motion is given by the equation: V(t) = 5 + 6 t n Determine the acceleration of the particle using: a(t) = dv/dt n

What’s next? MOVE IT!!

Description of Motion n n Motion is the apparent change in position of an object with time Use the following data to make a graph of distance versus time for a car traveling on a smooth, straight road. Time (s) Distance (m) 0 0 60 366 120 1219 180 2743 240 2896 300 2926 360 3962 420 5486 480 7163 540 7315 600 7315

Description of Motion How can this graph be used to describe the motion of the car? n What assumptions must be made? n What are the limitations of this graph? n

Description of Motion n n Determine the average velocity of the car from 360 to 540 seconds. If during this time interval the car was traveling through a school zone (speed limit 25 mph), what might be happening from 540 seconds on? A secant line may be used to determine the average velocity!

An amusing interlude n A policeman pulls over a woman and tells her she was going 60 miles per hour. The lady says, “That’s impossible, I’ve only been driving for fifteen minutes. ” The policeman responds, “ If you had looked at your speedometer, it would have read 60. ” The woman says, “My speedometer is broken, it always reads zero; does that mean I wasn’t moving at all? ”

Instantaneous Velocity n Determine the velocity of the car at 120 seconds 120, 1219 Did someone say, “tangent line”?

Tangent line method n n What factors limit the accuracy of this method? How is it possible to improve the accuracy of determining instantaneous velocity? A tangent line intersects the curve at only one point

PROBLEM ALERT II n Unless we know exactly how the woman was driving at 120 seconds, the limit is still an approximation n The best way to accurately determine the instantaneous velocity is if the equation for the graph is known.

Velocity as a derivative n Suppose we know that from zero to 180 seconds the woman drives with constant acceleration and her distance as a function of time is: x(t) = 0. 085 t 2 A derivative describes how one variable changes with respect to another variable n Velocity is the time rate of change of position The standard form for this equation is x=tn and n

Velocity as a derivative n Therefore the derivative of x=0. 085 t 2 is v=2(0. 085)t and the velocity at 120 seconds is v=2(0. 085)(120) or v = 20. 4 m/s n n A derivative is just another way to determine slope. Compare the derivative to the slope calculated using the tangent line method.