KINEMATICS

Kinematic concepts - Giancoli, Ch 2: 1-7 visit the Physics Classroom
1. Define displacement, velocity, speed and acceleration.
2. Define and explain the difference between instantaneous and average values of speed, velocity and acceleration.
· Displacement or change in position is the straight line length connecting the starting and ending points of motion, along with the direction of travel à displacement is a vector quantity
· Velocity describes how the position changes over time à velocity is a vector quantity, because it depends on displacement
· Average velocity is calculated as Ds / Dt (where s represents the displacement vector)
· Instantaneous velocity is defined as the velocity at a particular location -- in reality it is the average velocity over an indefinitely small interval of time (Dt approaches the value 0)
· Speed describes how the distance traveled changes over time à speed is a scalar quantity, not a vector quantity. Speed can be average or instantaneous, just like velocity.
· Acceleration describes how the velocity changes over time à acceleration is a vector quantity because it depends on the velocity.
· Average acceleration is calculated as Dv / Dt (where v represents the velocity vector)
· Instantaneous acceleration is defined as the acceleration at a particular location -- in reality it is the average acceleration over an indefinitely small interval of time (Dt approaches the value 0).
3. Describe an object's motion from more than one frame of reference.
· All motion is described from some point of view AND the description of the motion depends greatly on that point of view. The point of view is called a frame of reference since it defines the coordinate system (frame) in which motion is to be described.
· An example: an observer at the side of the road sees two cars traveling N at 20 m/s (car A) and 30 m/s (car B). What is the relative velocity of Car B from the point of view of (or relative to) car A? The motion of car A will change the observed velocity of car B -- Car B is still moving North, but from A's point of view it is only pulling away at 10 ms-1 (30 - 20). What is the velocity of Car A from the point of view of (or relative to) car B? The motion of Car B will change the observed velocity of car A -- Car A appears to moving at a velocity of -10 ms-1 (20 - 30). In other words, Car A appears to going South (-North) at 10 ms-1.
· To calculate the relative velocity of one object relative to a second object, subtract the velocity vector of the second object from the velocity vector of the first object.
Graphical representation of motion - Giancoli, Ch 2: 11 view a Java Applet
4. Draw and analyse distance–time graphs, displacement–time graphs, velocity–time graphs and acceleration–time graphs. Students should be able to sketch and label these graphs for various situations. They should also be able to write descriptions of the motions represented by such graphs.
· Forward (positive) motion is represented by a positive slope on a position-time graph. Backward (negative) motion is represented by a negative slope on a position-time graph.
· Constant velocity is represented by a linear, sloped position-time graph or a horizontal line on a velocity-time graph.
· Forward (positive) velocity is drawn in the first quadrant (above the time axis) on a v-t graph. Backward (negative) velocity is drawn in the fourth quadrant (below the time axis) on a v-t graph.
· Uniformly accelerated motion is represented by a linear, sloped v-t graph. Positive acceleration has a positive slope and negative acceleration has a negative slope.
· Uniformly accelerated motion is represented by a curved (parabola) position-time graph. Positive acceleration is an upward curved parabola. Negative acceleration is a downward curved parabola.
5. Analyze and calculate the slopes of displacement–time graphs and velocity– time graphs, and the areas under velocity–time graphs and acceleration–time graphs. Relate these to the relevant kinematic quantity.
· Slope of a position-time graph is the velocity of the motion. To find instantaneous velocity, find the slope of the line tangent to the curve at a specific location (for curved graphs).
· Slope of a velocity-time graph is the acceleration of the motion. To find the instantaneous acceleration, find the slope of the line tangent to the curve at a specific location (for curved graphs).
· The area between the velocity-time curve and the time axis represents the displacement of the motion.
· The area between the acceleration-time curve and the time axis represents the change in velocity for the motion.






Uniformly accelerated motion - Giancoli, Ch 2: 8-10
6. Determine the velocity and acceleration from simple timing situations.
Students should be able to interpret data from devices such as a light gate, strobe photograph or ticker timer. Analysis may involve graphing the data, taking measurements and applying kinematics concepts.



An example: the dots in the diagram represent successive positions of a car, taken at 1-second intervals, beginning with time = 0 seconds. 1 cm on the diagram represents 10 m. To find the velocity during any interval, you would measure the distance between successive dots, convert the measurement to meters, and divide by the elapsed time. To find the acceleration of the motion, you would need to find the average velocity for two successive intervals and then divide by the same time interval.
7. Derive the equations for uniformly accelerated motion.
· v = u + at this is the equation of a line on a v-t graph with u as the vertical intercept and a as the slope
· Ds = ut + 1/2at2 this is the area under a velocity time graph with u as the vertical intercept and a as the slope
· Ds = (u + v)t / 2 this is still the area under a velocity-time graph (u + v)/2 is the average velocity
· v2 = u2 + 2aDs solve the first equation for t, sub into the second equation and simplify
8. Describe the vertical motion of an object in a uniform gravitational field.
· Freefall motion is the motion of an object falling only under the influence of gravity. Air resistance is assumed negligible .
· Objects fall with uniform acceleration (9.8 ms-2 directed downward)
· All objects, regardless of their mass, fall with the same uniform acceleration.
· The same equations of motion can be used for freefall motion. If the upward direction is taken as positive motion, then a = - 9.8 ms-2 (the minus sign denotes a downward direction for the acceleration).
· The acceleration of an object moving vertically up or down is always 9.8 ms-2 directed downward, regardless of where in the trajectory the object is located.
9. Describe the effects of air resistance on falling objects.
· If air resistance is included in the problem, mathematically it becomes much more complicated. This is because the amount of air resistance changes in relation to the speed of the falling object.
· As objects fall, the amount of air resistance (directed upward) increases until it is exactly balanced by the downward pull of gravity. At this point, there is no net force on the falling object and it will fall with a constant speed from that point on. The constant speed that objects reach is called terminal velocity.
10. Solve problems involving uniformly accelerated motion.

· See Giancoli, Ch 2 Problems Level I: #2, 4, 15a, 19, 21, 35
Level II: #8, 11, 37, 41, 45, 61
Level III: #17a, 29, 51