(revised 08-08-03)

Horizontal and Vertical Motion

Speed = Change in position or distance traveled over time.

= A rate of change in position

= A scalar quantity – A quantity having a unit but with no reference point (non-directional).

Constant or uniform speed = 30 mi/hr, 60 mi/2 hr, 90 mi/3 hr

Velocity = Change in displacement over time.

= A rate of change in displacement.

= A vector quantity – A quantity having a unit with a reference point. A directional value is

implied.

Units: km/hr, m/sec, mi/hr, ft/sec

Instantaneous velocity = The velocity of a moving body at any one specific instant of time or at one specific

point in the path.

Terminal velocity = The point when a freely falling body ceases to accelerate due to the increasing effect

of air resistance. The point when the downward force due to gravity is balanced by

an equal upward force, caused by the tendency of air to resist the passage of the

body through it.

Acceleration = A change in velocity over time, or a time rate of change in velocity.

= A vector quantity.

= Due to an unbalanced force on a body, which follows Newton’s Second Law of Motion.

(F = m · a).

a = ΔV/t = (V_f – V_o)/t

Units = km/hr/sec, mi/hr/sec, m/sec/sec (m/sec²), ft/sec/sec (ft/sec²)

Graphic interpretation of acceleration:

A body moving with an unchanging speed over time is NOT accelerating.

ΔV/t = a positive linear slope, “a”, which means that uniform motion is present.

A body moving with a uniformly increasing speed over time produces a positive acceleration.

Graphical analysis of the illustration above provides the following information:

a = ΔV/t = (V₂ – V₁)/ t = (3 m/sec – 2 m/sec) / (9 sec – 6 sec) = (1 m/sec)/(3 sec) = 0.33 m/sec².

Interpretation: The body increases its speed 0.33 m/sec EACH SECOND of travel. In other words, during each second of travel, the body moves 0.33 m/sec faster than the second before.

a = ΔV/t = (V₂ – V₁)/ t = (2 m/sec – 6 m/sec) / (11 sec – 3 sec) = (- 4 m/sec)/(8sec) = - 0.50 m/sec²_.

Interpretation: The body decreases its speed 0.5 m/sec EACH SECOND of travel. In other words,

during each second of travel the body moves 0.5 m/sec slower than the second before.

In each of the two cases illustrated, the slope of the line is defined as the quotient of the ratio ΔV/t

(rise over run), and is known as acceleration, “a”. When ΔV is large, due to a large change in speed,

“a” is large.

Moving terms to create formulas:

ΔV/t = a

(V₂ – V₁)/ t = a

V₂ – V₁ = a · t

V₂ = V₁ + a · t or V_F = V_O + a · t

V = V_O + V_F / 2 substitute (V_F = V_O + a · t)

V = [Vo + (V_o + a · t)] / 2

V = (2Vo + a · t) / 2

V = Vo + ½ a · t

add the term, “t”:

V(t) = Vo(t) + ½ a · t (t)

V· t = Vo · t + ½ a · t²

Let V = s/t, so, s = V· t. Substitute “s” for V· t in the equation above.

S = Vo · t + ½ a · t²

S = Vo · t + ½ a · t² V_F = V_O + a · t

(V_F - V_O) / a = t

Substitute “t” in the equation, S = V_O · t + ½ a · t², with (V_F - V_O) / a .

This leads us to:

S = V_O · [(V_F - V_O) / a] + ½ a · [(V_F - Vo) / a ]²

S = (V_O·V_F – V_o² )/ a + ( V_F² - 2V_F ·V_O + V_O²) / 2a

S = (2V_O·V_F - 2V_O²) / 2a + (V_F² – 2V_F·V_O +V_O²) / 2a

S = (V_F² – V_O²) / 2a

V_F² = V_O² + 2·a·S