Motion of Objects

Much of what we study in physics involves the analysis of motion.

In order to describe the movement of an object, it is important to specify:

• The previous position of the object,
• Its current position, and
• Be able to predict where it could be at a later time.

This lesson will review early ideas about motion by Aristotle, Galileo, and Newton.
Next it will review the meanings and definitions of familiar words used in common vocabulary to describe motion.

In this study of motion, mechanics is the area of physics that involves the study of the motion of objects and the concepts of force and energy that cause the motion to change.

The two branches of mechanics are:

• Kinematics and

• Dynamics.

Kinematics is the branch of mechanics that focuses on describing the motion without investigating the forces that cause the motion. This description of motion generally involves the position and velocity of the object as a function of time.

Dynamics is the branch of mechanics, which deals with the study of the effects of forces (pushes and pulls) on the motion of an object.

Definitions of the Kinematical Variables

Vectors and Scalars

The motion of objects can be discussed in terms of physical quantities that describe the position of an object, how fast it is moving, and the time taken for the motion. In describing the motion of an object, it is usually necessary to note the initial and final positions of the object, thus determining the direction of the object's motion (such as to the right or to the left, due north or due east).

Quantities that can only be completely described in terms of the magnitude (or size) and direction are known as vectors.
Examples of vectors are acceleration, force, and electric field.

Those quantities that are completely described only in terms of the magnitude are known as scalars.
Examples of scalars are time, temperature, and energy.

Displacement

    A vector representing the change in the position of an object, is drawn from the initial to the final position.

    The most common units for displacement are meters (m), centimeters (cm), or feet (ft).

Average Speed

    The distance (a scalar quantity) an object moves in a given amount of time divided by the time.

    Units are (m/s), (cm/s), or (ft/s).

Average Velocity

    The displacement (a vector quantity) of an object divided by the elapsed time.

    Units are the same as average speed.

Instantaneous Velocity

    The rate of change of displacement.

    Units are (m/s), (cm/s), or (ft/s).

Instantaneous Speed

    The magnitude of the instantaneous velocity vector.

    Units are the same as instantaneous velocity.

Average Acceleration

    The change in the velocity (a vector quantity) of an object divided by the elapsed time.

    Units are (m/s²), (cm/s²), or (ft/s²).

Instantaneous Acceleration

    The rate of change of velocity.

    Units are (m/s²), (cm/s²), or (ft/s²).

Free Fall

    An idealized motion, in which air resistance is neglected and the acceleration is nearly constant.
    This acceleration is due to gravity and taken to be 9.80 m/s² or 32.2 ft/s² at the surface of the Earth.

The vector quantities have an obvious direction in space whereas the scalar quantities do not have a direction associated with them.

A vector is fully defined if its magnitude (size, including appropriate units) and direction are specified. The skill set derived from basic concept of vectors is used throughout the study of motion.

Review the relevant chapter and sections in your textbook to learn more about the outlined topics. Click on the hyperlinked words to read the related information online. You should be able to answer the questions below after covering the relevant material. Be sure to complete all online practice exercises to check your understanding before proceeding to the next topic.

1.      Distinguish between vectors and scalars.

2.      Displacement is the change in position of an object. It defines how far an object has moved from its starting position. Do Distance and Displacement represent the same quantity?

3.      What are the major differences between Speed and Velocity?

4.      Can you differentiate between Average speed and Instantaneous speed; Average velocity and Instantaneous velocity?

5.      Describe Acceleration and explain the difference between velocity and acceleration. How does the direction of the acceleration of an object affect the object's velocity? For an object moving at constant acceleration, how does the displacement of the object depend on the time taken to cover that displacement?

1.      Can you draw a vector diagram that depicts an object slowing down (decelerating) until it stops?

2.      Do the Motion Assignment question on describing motion using diagrams.

So far, we know that the motion of an object moving freely on earth is affected by the force of gravity. An object thrown by some force and continues to move on its own is a projectile. In the absence of the acceleration due to gravity, a projectile in free fall will move with a constant velocity. However, in the presence of gravity (assume acceleration, g = 10 m/s²) the motion of this projectile becomes complicated and its velocity changes from position to position along its path of motion! This acceleration due to gravity affects only the velocity component (v_y) along the vertical but has no effect on the horizontal velocity component (v_x). Therefore, we must consider the motion in two parts: the vertical and the horizontal. For consistent definitions, we shall adopt the convention that any vector quantity that is directed UPward is considered POSITIVE while any vector that points DOWNward is considered NEGATIVE!

Horizontally launched projectiles

Consider the motion of a ball rolled off the edge of a horizontal table surface and falls freely downward, or an arrow shot horizontally and falls freely as it moves toward the target at some distance away. After the ball leaves the edge of the table or the arrow is shot from the bow, no horizontal forces act on the ball or arrow. One observes the curved path of motion as the object falls vertically downward and horizontally away from its starting position.

The vertical component of its velocity is perpendicular to the ground under the influence of the force of gravity, just as we have already learned for vertical motion under gravity. As the object starts to fall from a position of rest (at y = 0), the downward (vertical) velocity increases by 10 m/s for every second of its fall due to gravitational acceleration. Also, the object falls downward and covers successively greater vertical distances along the y­-axis (such that the vertical displacement Δy = - d) for equal intervals of time according to the equation for distance fallen: d = ½gt².

Using g = 10 m/s², the vertical distance of fall at time t, can be found as: d = 5t².

Ignoring the effects of air resistance to the object’s motion and the rotation of the earth, there is no horizontal acceleration on the object. So, its horizontal velocity component does not change and remains constant during its free fall motion. Thus, this projectile covers equal horizontal distances in equal intervals of time.

This horizontal distance is equal to the horizontal component of the velocity multiplied by the time of fall, i.e. x = (v_x)t.

The diagram above shows blue “ghost” images of the actual positions of the ball during the free fall.  The horizontal green arrows (of equal lengths) show that the ball moves equal horizontal distances every second of its motion. The red images show the successive vertical positions of the ball every second of its motion, as if it were falling vertically downward in free fall. By comparing the blue and red images, notice that the actual vertical positions (blue images) match the predicted vertical positions (red images) for every second during the free fall.

We can conclude that the curved path (parabolic trajectory) of the object is a combination of the horizontal and vertical motions.

Upwardly launched projectiles

What about an object thrown initially at an upward angle? Typical examples include a  golf ball hit from the tee, a leaping frog, water sprayed from a hose or sprinkler, a football thrown to a receiver, or an athlete doing a long jump.

Consider the motion of a football thrown at an upward angle θ, measured from the ground. Since the motions along the vertical and horizontal directions are independent of each other, you must find the vertical component (v_oy) and horizontal component (v_ox) of its initial velocity. These components can be found using the trigonometric ratios according to the definitions:

Horizontally: v_ox = v_o cos θ, and
vertically: v_oy = v_o sin θ.

From the figure shown below, notice that horizontally the ball (as depicted by the green ghost images) moves at a constant velocity component v_ox, thereby covering equal distances in equal time intervals (i.e. per second in this diagram). You should expect this because there is no horizontal acceleration!

However, along the vertical direction, the red ghost images of the ball show the path of the ball as if it were thrown vertically upward with an initial velocity component v_oy. Due to the downward-directed force of gravity, the ball covers successively shorter distances in equal intervals of time as it climbs to the highest point of its vertical motion before falling downwards. Thus, the combination of the vertical and horizontal motions of the ball occur simultaneously (at the same times) as shown by the path of the blue ghost images.

At the maximum height (i.e. at the highest vertical point) of the motion, notice that the ball still has the horizontal velocity component even though the vertical component of the velocity at that instant is zero! Beyond this maximum height, as the ball starts falling, under the influence of the force of gravity still directed downwards toward the ground, the vertical velocity component of the ball increases from zero (at the maximum height). Thus the falling ball now covers increasingly greater distances in equal time intervals.

Do not forget that the horizontal velocity component still remains constant since there is no horizontal acceleration!

In addition, since the constant acceleration due to gravity “slows” the upward motion and “speeds” the downward motion at the same rate, the amount of time it takes the projectile to rise to its maximum height is equal to the time it takes to fall back to its starting position, as long as it moves freely.

Therefore, with a constant horizontal velocity and a varying vertical velocity, the total velocity of the ball along its path changes from point to point as the ball rises and falls. The magnitude of this (combined) total velocity (v) at any point along the trajectory is calculated using vector methods (Pythagorean Theorem):
 v² = v_ox² + v_oy².

Figuring the distances moved during the projectile motion

Horizontal displacement

The horizontal distance from its starting position moved by a projectile in free fall under gravity is equal to the horizontal component of the velocity multiplied by the time of fall. We generalize this relationship as:
  Δx = (v_ox)t.

Vertical displacement

Recall that the acceleration due to gravity (g = 10 m/s²) causes the projectile to begin falling toward the earth as soon as the projectile is released to fall freely. During this “fall to the earth”, the projectile covers successively greater vertical distances downward along the y­-axis (such that Δy = - d) for equal intervals of time according to the equation: d = ½gt².

For horizontally-launched projectiles, the initial vertical velocity component v_oy = 0 (zero). For upwardly or downwardly launched projectiles, the
initial vertical velocity component, v_oy = v_o sin θ, while the
(constant) horizontal velocity component is v_ox = v_o cos θ.

However, we must remember that the projectile must cover some initial displacement along the vertical (y) axis due to its initial vertical velocity component v_oy. This vertical displacement is (v_oy)t.

This vertical displacement is initially upward and positive if v_oy points vertically upward (positive). On the other hand, the displacement may be initially downward and negative if v_oy points vertically downward (negative).

Adding these two vertical displacements, the total vertical displacement is given as:

 Δy = (v_oy)t - ½gt²