Introduction to Uniform and Accelerated Motion
This page originally compiled Feb 2007 by TFR
Motion is a change of position of a body relative to something. The something, for our purposes, is any reference against which Newton's laws of motion are valid, known as an "inertial reference frame". This simply means that no outside accelerations that we don't know about are present to cause strange things to happen, for instance things sliding across the dash of a car as the car makes a sudden turn. If we did not know what was happening, this would seem, to us inside the car, to be magic. As long as we keep the definition in mind, everything will work. Here we will generally be defining motion of our rocket relative to the "stationary" Earth.
Linear motion is motion in a straight line, also called motion of translation.
Angular motion is motion in a circle about an axis, also called motion of rotation.
Uniform motion is motion which does not change its speed or direction, that is, it is not accelerated.
Accelerated motion is motion which does change its speed or direction, generally in response to a force.
Displacement (s) is an amount of motion, for instance 20 feet or 34 kilometers.
s = vt
Examples:
A car moving at 20 km/hr moves how far in 3 hours?
s = 20 × 3
s = 60 km
An airplane is moving at 550 mi/hr. How far does it travel in 5 seconds?
Convert v to miles/second: v = 550/3600 (the number of seconds in 1 hour)
s = 0.153 × 5
s = 0.765 miles
Speed is the rate of motion, equal to distance/time, for instance, miles per hour or km per second.
Velocity (v) is a directed speed, a vector quantity. Some simple examples are: 500 miles/hour due North, 1 km/second straight up. The terms velocity and speed are sometimes interchanged when the direction is not important to the outcome, for instance a falling object moves straight down, so describing its velocity as 100 feet/second is not ambiguous.
v = s/t
Examples:
A car moves 60 km in 3 hours. How fast is it going?
v = 60 / 3
v = 20 km/hr
A balloon rises 1000 feet in 40 seconds. What is its vertical speed in mi/hr?
v = 1000 / 40
v = 25 ft/sec
Convert to ft/hr: v = 25 × 3600
Convert to mi/hr: v = 90000/5280 = 17.0 mi/hr
Acceleration (a) is the rate of change of velocity, equal to velocity/time, for instance 32 feet/sec per second, written 32 ft/sec2. Again, acceleration is a vector quantity and again, when context renders this unimportant we can ignore its direction.
a = (v - v0)/t, where v0 is the starting velocity
a = vc/t, where vc is the change in velocity, (v - v0)
vc = at
Examples:
An object's velocity increases from 20 ft/sec to 30 ft/sec in 15 seconds of time. What is the acceleration?
a = (30 - 20)/15
a = 0.667 ft/sec2
An object is accelerated at 10 cm/sec2 for 5.3 seconds. How much faster is it going?
vc = 10 × 5.3
vc = 53 cm/sec faster
An object is accelerated at -62 ft/sec2 (i.e. it is decelerated) for 2 seconds. What is its change in velocity?
vc = -62 × 2
vc = -124 ft/sec, i.e. 124 ft/sec slower
Uniformly accelerated motion occurs when the velocity is changing at a constant rate. This means the acceleration remains the same from time to time, for instance, during the 1st second our object accelerates from 0 to 10 ft/sec, during the 2nd second, from 10 ft/sec to 20 ft/sec, during the 3rd, 20 ft/sec to 30 ft/sec, etc. It is apparent that the acceleration is a constant 10 ft/sec2 throughout these 3 seconds. Under these conditions, it is easy to determine the final velocity:
vf = v0 + vc, and, since vc = at
vf = v0 + at
Now, since the acceleration is uniform, we can say that the average velocity from start time to final time is:
va = (v0 + vf)/2
Substituting from above,
va = (v0 + vf)/2 = (v0 + (v0 + at))/2 = (2v0 + at)/2
va = v0 + 1/2at,
giving us an expression for average velocity in terms of starting velocity, acceleration, and time. With this, we can calculate the displacement, which is simply the average velocity × time, as above:
s = vat
Substituting from above,
s = (v0 + 1/2at)t
s = v0t + 1/2at2,
giving us an expression for displacement in terms of starting velocity, acceleration, and time.
In some cases, we might wish to calculate without reference to time. If we solve vf = v0 + at for time:
t = (vf - v0)/a, and, since va = (v0 + vf)/2, and s = vat, as above:
Substituting, s = [ (v0 + vf)/2 ] × [ (vf - v0)/a ] = [ v0vf - v02 + vf2 - vfv0 ] / 2a
s = (vf2 - v02) / 2a
Examples:
From our discussion above,
vf = 0 + 10 × 3 = 30 ft/sec
va = 0 + 1/2 × 10 × 3 = 15 ft/sec
s = 0 × 3 + 1/2 × 10 × 32 = 45 ft
s = (302 - 02) / (2 × 10) = 45 ft
We now have some powerful tools to analyze some of the motions which interest us.
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Last Revised: Feb 2007