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Study Of Motion

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A. Describing, Measuring, and Displaying Motion Scenarios: In everyday conversation, as well as in physics, math, and engineering, the motion of objects are discussed. The most common discussions focus on either constant speed or acceleration scenarios. In everyday conversation usually a particular object is being discussed ranging from a skate board to an automobile or some other form of motorized transportation. In the sciences and in engineering expressing information about constant speed, velocity (not exactly the same as speed), and acceleration are concepts that are quantified and need to be measured, recorded, and displayed. Two common methods of displaying such observations (generally called data) is through the use of ticker tape measurements and vectors. You may click on either of the two links below to obtain more information about these methods of expressing data about motion.

1. Motion described using "Ticker Tape"

2. Motion described using "Vectors"

B. "Motion in One Dimension"  Notes on motion and solving problems involving motion.

1. Displaying Problem Solving Information (including variable symbols)

Displaying Problem Solving Information
1. All problem solving solutions use a five step problem solving format consisting of a(n):

 Given statement, G:      Find statement, F:      Equation statement, E:      Substitution statement, S:      Answer statement, A:

2. A bar over a variable symbol is used to indicate an average value.

3. A subscript of aver, or ave enclosed in parentheses, (ave), following a variable symbol could also be used.

4. A scalar quantity variable is printed in physics text books in non bold type.

5. A vector quantity variable is printed in physics text books in bold type, sometimes with an arrow above the symbol.

2. Constant Speed and Velocity

Constant Speed is the measure of the rate of "how fast" an object is moving without concern for the direction in which the object is traveling.

Speed belongs to a category of measurements called scalar quantities. A scalar quantity only answers the quantitative question "how much?".

Constant Velocity is a measure of both an object's speed and the direction in which the object is traveling.

Velocity belongs to a category of measurement called vector quantities. A vector quantity answers two questions, which are "how much?" and "in what direction?". Direction is most often expressed as an angle measured from some reference point.

--> Example of Average Speed

 A bicycle rider travels 22 km in 3.25 hours. What is the average speed of the bicycle rider? The magnitude of v(ave) is average speed G: distance (d) = 22 km;  time interval (t) = 3.25 hours F: Average speed, v(ave) E: v(ave) = d / t S: v(ave) = 22 km / 3.25 h A: v(ave) = 6.8 km / h

--> Example of calculating Displacement during Constant Velocity

Displacement is the distance (a scalar measurement) an object has moved and the direction the object has moved. It is a vector measurement. Displacement is also called "change in position".

 A vehicle moves along at 60 km / h for a time interval of 30 minutes due North (directly North). What is the displacement of the vehicle? G: v(ave) = 60 Km / h, t = 30 minutes; Note 30 minutes is 1/2 hour or 0.5 h F: Displacement, d E: d = v(ave) t S: d = [60 km / h] [0.5 h] A: d = 30 km

3. Uniform (constant) Acceleration

Acceleration is also a vector quantity and represents a measure of the rate of change in the velocity of an object. Variable symbols representing vector quantity measurements can be written with an arrow placed over the variable symbol or in a typed font printed in bold type.

--> Example of Average Velocity

 A high speed train takes 2.00 hours to go from Paris 454 km due south to Lyons. What is the average velocity of the train? G: displacement (d) = 454 km, S;  time interval (t) = 2.00 hr F: Average velocity, v E: v = d / t S: v = 454 km, S / 2.00 hr A: v = 227 km / hr, S

--> Example of Acceleration

Acceleration is the change in velocity divided by the interval of time in which the change occurs.  The unit of acceleration is m/s^2 or km/h^2, etc.

 The velocity of an automobile increases from 0 to 14 m/s, E, in 3.5 s.  What is its acceleration? G: initial velocity v(i) = 0 m/s ; final velocity v(f) = 14 m/s, E ; time interval (t) = 3.5 s F: Acceleration (a) E: a = [v(f) - v(i)] / t S: a = (14 m/s  0 m/s) / 3.5 s A: a = 4.0 m/s^2, E

--> Example of Negative Acceleration

 An automobile slows from 14.0 m/s, E, to 7.0 m/s, E, in 20 s.  Find its acceleration? G: intial velocity v(i) = 14.0 m/s, E ; final velocity v(f) = 7.0 m/s, E ; time interval (t) = 2.0 s F: Acceleration (a) E: a = [v(f) - v(i)] / t S: a = (7.0 m/s  14 m/s) / 2.0 s A: a = -3.5 m/s^2, E

4. Other variations of Uniform Acceleration Problems

Final Velocity can be found from the initial velocity, acceleration, and time interval.

--> Example of calculating Final Velocity after uniform acceleration from initial velocity, acceleration, and time

 A ball rolling down an incline for 5.0 s undergoes a uniform acceleration of 4.2 m/s^2.  If the ball has an initial velocity of 2.0 m/s when it starts down the incline, what is its final velocity? G: time (t) = 5.0 s ; acceleration (a) = 4.2 m/s^2 ; initial velocity v(i) = 2.0 m/s forward down the incline F: final velocity v(f) E: v(f) = v(i) + at S: v(f) = 2.0 m/s + 4.2 m/s^2 (5.0 s) A: v(f) = 23 m/s, forward down the incline

The average velocity of a uniformly accelerating object is the middle velocity as in v(ave) = [v(f)+ v(i)] / 2

Remember that Displacement is the distance (a scalar measurement) an object has moved and the direction the object has moved. It is a vector measurement. Displacement is also called "change in position".

The displacement of an object can be found from its initial and final velocities and the time interval over which the change occurs.

--> Example of calculating Displacement during Uniform Acceleration from final velocity, initial velocity, and time

 What is the displacement of a train as it is accelerated uniformly from 22 m/s to 44 m/s, west in a 20.0 s time interval? G: initial velocity v(i) = 22 m/s, W ; final velocity v(f) = 44 m/s, W ; time interval (t) = 20.0 s F: displacement, d E: d = [ ( v(f) + v(i) ) / 2] t S: d = [ (44 m/s + 22 m/s) / 2 ] (20.0 s) A: d = 660 m, W

The displacement can also be found from the initial velocity, acceleration, and time interval

--> Example of calculating Displacement during Uniform Acceleration from initial Velocity, Acceleration, and Time

 A car starting from rest is accelerated at 6.1 m/s^2, S.  What is the cars displacement during the first 7.0 s of acceleration? G:  intial velocity (vi) = 0 m/s ; acceleration (a) = 6.1 m/s^2 ; time interval (t) = 7.0 s F:  displacement (d) E:  d = d(i) + v(i) t + 0.5 a t^2 S:  d  = 0 m + 0 m/s (7.0 s) + 0.5 (6.1 m/s^2) (7.0 s)^2 A:  d = 150 m, S

Final velocity can also be found from the initial velocity, acceleration, and distance traveled.  And Acceleration can be found from the initial velocity, final velocity, and the distance traveled.

--> Example of calculating Acceleration during Uniform Acceleration from final velocity, intial velocity, and displacement

 An airplane must achieve a velocity of +71 m/s for takeoff.  If the runway is 1.0 x 10^3 m long, what must the acceleration be? G: initial velocity (vi) = 0 m/s ; final velocity (vf) = +71 m/s ; displacement (d) = 1,000 m F: acceleration (a) E: a = [ v(f)^2  v(i)^2 ] / [ 2 d ] S: a = [ (71 m/s)^2  (0 m/s)^2 ] / [ 2 x 1000 m ] A: +2.5 m/s^2

The displacement is the same if the initial and final velocities are exchanged and acceleration is changed to deceleration

--> Example of finding Displacement when Velocities and Times are known

 Determine the displacement during constant (uniform) acceleration of an airplane that is accelerated from +66 m/s to +88 m/s in 12.0 s. G:  initial velocity v(i) = +66.0 m/s ; final velocity v(f) = +88.0 m/s ; time (t) = 12.0 s. F:  displacement, d E:  d = [ ( v(f) + v(i) ) / 2 ] t S:  d = [ (88.0 m/s + 66.0 m/s) / 2 ] [ 12 s ] A:  d = +924 m

Acceleration due to gravity problems requires the use of the same basic equations, but the acceleration symbol a is changed to a g for the acceleration due to gravity.

--> Example of a falling body (object) problem

 A 20 kg object is suspended 100 m in the air by a large construction crane.  The mass is released and allowed to fall to the ground. a. How long will it take for the object to reach the ground? b. What is the final velocity of the object just before it strikes the ground? Solution to part a: The equation d = v(i) t + 0.5 g t^2 allows you to solve for the time. Since the initial velocity for an object that is dropped is zero, the vi t in the equation will drop out.  The d = 0.5 g t^2 portion of the equation is used to solve for the time. E: t = (2 d / g )^0.5 S: t = [ 2 (100 m) / 9.8 m/s^2 ]^0.5 A: t = 4.52 s Solution to part b: Note: The acceleration downwards is the acceleration due to gravity, so a = g. E: v(f) = v(i) + a t S: v(f) = 0 m/s + 9.8 m/s^2 (4.52 s) A: v(f) = 44.3 m/s, downwards It would be proper to express the answer as a -44.3 m/s, where the negative in front of the number implies downwards. The use of the negative for the downwards direction would be valid in situations defined by you or others where up was defined as being the positive direction and down was defined as being the downward direction.