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II. Motion Two branches of Newtonian mechanics1: Kinematics:
describes motion without consideration of the causes of motion.
Dynamics: concerned with the study of forces and torques2 and their
effect on motion. Definition of Concepts Related to the Study of
Motion: Time: Physicists define time as simply a measure of change.
If nothing changes then it is impossible to measure time.
Displacement: Displacement is the shortest distance from the
initial to the final position of a point. It is the length of an
imaginary straight path, typically distinct from the path actually
travelled. Displacement, unlike distance, is a vector3. Velocity:
Velocity is the change in displacement divided by the change in
time. Velocity is a vector and thus has magnitude4 and direction.
The average velocity (V) is equal to a change in displacement5
divided by change in time:
or, for convenience, we will use (change in distance) instead of
(change in displacement): Equation 2.1
And whenever there is change, there is always a difference:
1 Newtonian mechanics is also known also known as classical
mechanics, a field in physics concerned
with the laws of motion. 2 Forces and torques (forces that cause
a rotation) are the main causes of motion. Therefore, dynamics
is
mostly concerned with these two. 3 A scalar is a quantity that
has magnitude but no direction. A vector is a quantity that has
both magnitude
and direction. 4 The absolute value of the magnitude of velocity
is called the speed, which is a scalar.
5 Remember, displacement is a vector.
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Therefore,
Consider for example a car which travelled 6.0 km in 15 min.
What is the average velocity of this car? Solution. Using your
equation 2.1:
Converting that to your standard SI unit for speed ( ) would
yield:
(
) (
)
The average velocity V of the car is 6.7
.
Acceleration: Acceleration is the change in velocity divided by
the change in time.
Because velocity is a vector, acceleration is also a vector.
The average acceleration is defined by:
Equation 2.2
We have mentioned that whenever there is change, there is always
a difference in the
initial and the final values:
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Therefore,
Consider for example a car, starting from rest (0
) and accelerated to 90
in the first
15 seconds to the East. What is the acceleration of this
car?
Solution.
Converting that to the standard SI unit for acceleration (
) yields:
(
) (
)
MOTION EQUATIONS AND PROBLEM SOLVING
It is possible to solve problems by deriving equations from your
common sense,
intuition, and a little bit of algebraic manipulation.
We know that
Equation 2.3
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And if we assume that and , then
and
Let these be your
Equation 2.4
Now using your equation 2.1 and assuming 2.4, we can deduce that
the average
velocity is simply displacement6 over time:
Now manipulate the resulting equation
Equation 2.4.a
The average velocity is given by
Equation 2.4.b
6 Remember we treat (change in distance) as (change in
displacement). This is technically incorrect because
distance is a scalar while displacement is a vector. However, we
mentioned that we replaced the symbols for convenience
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Replace the in equation 2.4.a with the value of in 2.4.b (that
is,
), we derive
Equation 2.5
(
)
Consider for example a car running at
and covered a distance of 1500.0
meters. How much time will it take for the car to travel this
distance at that velocity
(assuming velocity is constant, i. e. zero acceleration).
Solution. By manipulating equation 2.5 we deduce that
(
)
( )
Plugging in the numbers,
( )
We can also combine equations 2.2 and 2.3 (assuming equation 2.4
is the case):
( )
and now assuming ,
( )
-
( )
Or rearranging that, we obtain
Equation 2.6
Equation 2.6 is useful in cases where you want to know the final
velocity of an object
but you dont know the distance traveled but you have knowledge
of its initial velocity,
acceleration and the time it takes for that acceleration to
happen.
Consider for example a runner from rest (
) accelerating to
for the first 0.75
seconds. What is the final velocity of this runner?
Solution. Use equation 2.6:
(
) ( )
(
) ( )
Suppose for example that the same runner began decelerating
to
we can calculate
the time it takes for this runner to stop. This can be done by
algebraically manipulating
equation 2.6:
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The final velocity is
since at that point the runner would have stopped. The
acceleration is
since the runner is decelerating (accelerating in the
opposite
direction7) Then using the equation,
We can also combine equations 2.5 and 2.6 to obtain another
equation. Equation 2.5 is
(
)
We can use equation 2.6
to replace the in equation 2.5 with :
( ( )
)
(Note: this is to eliminate the final velocity in the equation
so that even if you dont
know the final velocity you can still solve other problems.)
( ( )
)
( )
)
7 Remember, acceleration is a vector!
-
Equation 2.7
Consider for the previous example. The runner has an initial
velocity of
and
decelerated to
within 3 seconds. What is the distance covered by this runner
after 3
seconds?
Solution. Note that the acceleration is
because deceleration is acceleration in the
opposite direction:
(
) ( )
(
) ( )
( )
(
) ( )
( )
( )
( ) ( )
We can also combine equations 2.5 and 2.6 to obtain another
equation that will
eliminate time instead of velocity. Equation 2.5 is
(
)
We can use equation 2.6
and rearrange that to obtain
-
Now replace the in equation 2.5 with
:
(
)
(
) (
)
(
)
(
)
(
)
Equation 2.8
Equation 2.8 is useful in finding velocities if we dont know the
time but know the initial
velocity, the displacement covered, and the acceleration.
Consider for example a car accelerating from rest to
and covered a distance of
400 meters (assuming the acceleration is constant). What is the
velocity of this car?
Solution. Use equation 2.8:
(
)
(
) ( )
(
)
(
) ( )
(
) (
)
(
) (
)
-
(
)
Summary of Motion Equations:
Equation 1:
Equation 2:
Equation 5: (assuming and as in equations 3 and 4) and
(
) (when there are two velocities)
Equation 6:
Equation 7:
Equation 8:
(Note: s is displacement which is simply distance with direction
in accordance to the
Cartesian coordinate)
