Sections
In 1-d kinematics, we are concerned with motion in a single straight line which we ordinarily think of as the x-axis or y-axis of a Cartesian coordinate system.
Note that t**p is a power of variable t.
From calculus, the derivative of function f(t)=t**p where t is the independent variable (and we are thinking of t being time in this lecture) is taken thus:
f(t) = t**p df(t)/dx = pt**(p-1) except df(x)/dt = 0 if p=0.Proof for integer p ≥ 1 :
df(t)/dx = limit_(h→0) [f(t+h)-f(x)]/h
= limit_(h→0) [(t+h)**p - t**p]/h
= limit_(h→0) [t**p+pt**(p-1)*h + ... - t**p]/h
= limit_(h→0) [pt**(p-1)*h + ...]/h
= pt**(p-1) , QED
where we have used the binomial theorem.
If p = 0 , then f(t) = 1 which clearly has zero
derivative.
For other cases of p,
we need more machinery which leave
sine die.
Behold:
df(t)/dt = t**p then f(t) = t**(p+1)/(p+1) + t_0 for p = ≠ -1 f(t) = ln(t) + t_0 for p = -1 ,where t_0 is constant of integration and where we have just reversed the process of differentiation for p = ≠ -1 and for p = 1, we need more machinery which leave sine die---in this case, we get natural logarithm of t, ln(t)
Given constant acceleration "a", we find the first two antiderivatives:
a constant acceleration
v = at + v_0 velocity
x = (1/2)at**2 + v_0*t + x_0 displacement
where v_0 and x_0 are
constants of integration
which in this case we identify as
time-zero constants:
i.e., initial conditions.
Prove the above equations for yourself.
Right now.
They are all vectors which in loose physics jargon means they are quantities with both magnitude and direction.
In one dimension, there is only positive and negative directions.
Note for 1-dimensional cases:
Thus,
Δ x > 0 is a positive displacement.
Δ x = 0 zero
displacement.
Δ x < 0 is a negative displacement.
The upper-case Greek letter Delta Δ is the common symbol for "change in".
The magnitude of displacement is distance.
Velocity can be positive, zero, or negative.
The magnitude of velocity is speed.
But note often people---including yours truly---will use velocity to mean both velocity and speed. Context decides what is meant.
Acceleration can be positive, zero, or negative.
The magnitude of acceleration is just called acceleration too---this is a sort of defect of the terminology.
If an acceleration points opposite to velocity (i.e., has a different sign), it's called a deceleration.
(1) x = (1/2)at**2 + v_0*t + x_0 displacement
(2) v = at + v_0 velocity
Absolutely, positively, they do NOT apply exactly when
acceleration is
NOT constant.
The 2 equations are algebriacally independent---one CANNOT be derived from the other by algebra.
How many variables are contained in the 2 equations? Count them. What are they?
Answer: 6. The are x, a, t, v_0, x_0, and v.
How many unknowns can we solve for?
Answer: 2.
In general, one can only solve for n equations for n unknowns. If one has n+1 unknowns, a solution is NOT in general possible.
So in general, for constant-acceleration problems you must be given 4 knowns and can solve from the 2 constant-acceleration equations.
The solution is by elementary algebra.
You can speed up the solution for constant-acceleration problems by 3 other constant-acceleration equations generated from the first 2 by algebra.
These 3 other equations are NOT algebraically independent of the first 2, and so you CANNOT solve more unknowns than 2, but can reduce the solution to solving 1 equation in all cases.
Eliminate t from equation (1) using equation (2) to obtain an equation for v**2. Do it.
Eliminate a from equation (1) using equation (2) to obtain a new equation for x. Do it.
Eliminate v_0 from equation (1) using equation (2) to obtain another new equation for x. Do it.
The set of 5 constant-acceleration equations:
Number Equation Missing Variables Name?
(1) x = (1/2)at**2 + v_0*t + x_0 v
(2) v = at + v_0 x,x_0
(3) v**2 = v_0**2 + 2a(x-x_0) t The timeless equation
(4) x = (1/2)*(v_0+v)t + x_0 a
(5) x = -(1/2)at**2 + v*t + x_0 v_0
Equations (1), (2), and (3) (i.e., the timeless equation) get the most use in problems
and equation (5) rarely turns up just because people seldom write problems that make use of it.
First identify the knowns and unknowns.
Some of them will be camelflaged. Often seeing through the camelflage is the whole problem.
You must be given 4 knowns.
To find one unknown out of two unknowns, you solve the equation that does NOT contain the unknown you do NOT want.
So the problem is reduced to solving for one unknown with one equation.
There are two common ways of complicating constant-acceleration problems:
In this case, the final conditions of one phase are initial conditions of the next.
Well many things can happen and many simple problems are possible.
The multiple-phase problems discussed in section Constant Acceleration are a kind of non-constant acceleration problem.
Average velocity, average acceleration, constant jerk, approximately constant acceleration.?????