# Describing Motion with Equations There are a variety of quantities associated with the motion of objects – displacement (and distance), velocity (and speed),

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Describing Motion with Equations There are a variety of quantities associated with the motion of objects – displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motiondisplacementvelocity acceleration

The kinematic equations of motion are derived under the assumption of constant acceleration. While this may at first seem to be a restriction, there are a large number of problems where the acceleration is a constant.

Summarizing, these four kinematic equations of motion are written as The first three equations relate displacement, velocity, and acceleration in terms of time while the fourth equation does not contain the time.

2-1 A train starts from rest (at position zero) and moves with constant acceleration. On first observation the velocity is 20 m/s and 80 s later the velocity is 60 m/s. At 80 s, calculate the position, average velocity, and the constant acceleration over the interval.

Calculate the distance traveled over this 80 s: The average velocity is If the acceleration is constant then the average velocity is the average of 20 m/s and 60 m/s, or 40 m/s, and at an average velocity of 40 m/s and 80 s, the distance traveled is 3200 m.

2-2 For the situation of problem 2- 1, calculate the position of the train at 20 s.

2-3 For the situation of problem 2-1 find the time required for the train to reach 100 m

2-4 For the situation of problem 2-1 find the velocity of the train at 120 m.

It is also important to remember that each of these terms have Have specific meanings. a. v 0 initial velocity b. v final velocity c. d total displacement d. d 0 initial displacement e. a acceleration from a moving object or gravity f. t final time g. t 0 initial time There are keys to remember, a.Starting from rest 0m/s b.Coming to rest 0m/s c.Coming to rest after _________m d.Total time of trip was _______

The kinematic equations were developed by Sir Isaac Newton, several centuries ago. They actually relate to Calculus. The four kinematic equations are: The first two equations are used the most.

The next two equations are:

Learn to use the equations to determine unknown information about an object's motion. include the following steps to do so: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation which will be used to determine the unknown information from the known variables. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown. 6. Check your answer to ensure that it is reasonable and mathematically correct.

The application of these four equations to the motion of an object in free fall can be aided by a proper understanding of the conceptual characteristics of free fall motion. These concepts are as follows: 1. An object in free fall experiences an acceleration of –9.8 m/s/s. (The negative (–) sign indicates a downward acceleration.) Whether explicitly stated or not, in the kinematic equations the acceleration for any freely falling object is ALWAYS –9.8 m/s/s. 2. If an object is dropped (as opposed to being thrown) from an elevated height to the ground below, the initial velocity of the object is 0 m/s.

3. If an object is projected upwards in a vertical direction, it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. 4. If an object is projected upwards in a vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of –30 m/s when it returns to that same height.

Derivation of Algebraic Derivation: Since the acceleration is constant, the average acceleration is also the same as the instantaneous acceleration. From the definition of acceleration we havedefinition of acceleration Multiplying both sides by t and solving for v. Thus we seethey are equivalent when the acceleration is constant.

Calculus Derivation: Start with the definition of instantaneous acceleration and multiply both sides by the differential dt.instantaneous acceleration Next integrate both sides. Since the acceleration is constant and can be taken out from inside the integral over time

Derivation of Analytical Geometry Derivation: In analytical geometry it can be shown that the average value of a straight line is the arithmetic average of its end points. The plot of the velocity versus time of an object undergoing a constant acceleration is a straight line.

Given the definition of average velocity, and the fact that the average velocity is the arithmetic average of the initial and final velocities, we find The average velocity can only be expressed as an arithmetic average of initial and final velocities when the acceleration is constant. It is not true in general.

Note that the average location of an accelerating object is not half way between the starting and ending locations. An objects spends a lot more time in the first half of its distance interval than in the last half because it is going slower in the first half than in the last half. However, if the object is moving at a constant speed, then the average location is at the midpoint of its motion since the object's location is a linear function of time.

Calculus Derivation:For constant acceleration We can evaluate the time average of the velocity directly since we know the velocity function's time dependence.

Derivation of Algebraic Derivation: Starting with and substituting we find

Calculus Derivation: Start with the definition of instantaneous velocity. Next multiply both sides by the differential dt and integrate

Derivation of Start with Solve for time and substitute it into the distance equation to eliminate the time.

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