Kinematics in Physics
These are my notes and worked examples of kinematics in physics.
The motion of a particle is described by giving position, velocity, and acceleration, usually as a function of time. We start with one dimension. Distance means total distance traveled, while displacement is the actual difference between endpoint and the beginning.
\[Velocity = \frac{displacement}{time}\]
and:
\[Acceleration = \frac{v2v1}{time}\]
This velocity calculation fits our present definition and is equivalent to drawing a straight line between points on the smooth curve of x versus t. Calculating velocity this way presents a problem, because, depending on the interval, we will get different answers for the velocity. To find the velocity at t=1, we found x at t=1 and x at t=2 and performed the velocity calculation. This is not a good approximation to the velocity at t=1 because the average is between t=1 and t=2. For better approximations, shorten the time interval centered about t=1. Getting it as small as you can will give better results. So, the slope of the straight line tangent to the curve at x=1 represents the velocity at x=1.
Instantaneous Velocity and Acceleration
A more versatile definition of velocity is \(\frac{\Delta x}{\Delta t}\), where the interval \(\Delta t\) is very small and centered about the time where the velocity is desired. This approach leads to a general method for obtaining an expression for velocity that can be evaluated at any point rather than going through the numeric calculation whenever we want a velocity. This definition is stated as the instantaneous velocity.
\[v = \lim\limits_{\Delta t \to \0} = \frac{dx}{dt}\]
This definition of the derivative is the slope of the tangent to the curve evaluated at the point in question. If we want to find the velocity of any particle traveling according to a polynomial relationship between x and t, all we need is a general technique for finding the slope of the tangent to the polynomial at any point.
The instantaneous acceleration is defined as the value of \(\Delta v / \Delta t\) as \(\Delta t\) approaches zero.
\[a = \lim\limits_{\Delta t \to \0} = \frac{dy}{dt}\]
The instantaneous acceleration is the slope of the tangent to the curve of v versus t. The general expression for the slope of any polynomial of the form \(x = ct^{n}\). The expression for the slope at any point is \(cnt^{n1}\). Stating this in calculus terms, for any function \(x = ct^{n}\), the derivative of the function is \(cnt^{n1}\). This can be verified in the case of the parabola by taking successively smaller intervals of \(\Delta t\) and \(\Delta x\) at any point t to verify that the slope at any point is 2t.
Kinematic Equations of Motion
The kinematic equations of motion are derived under the assumption of constant acceleration. While this may seem at first to be a restriction, there are a large number of problems where the acceleration is constant. The simplest and most obvious are falling body problems, or problems involving bodies falling on or near the surface of the earth where the acceleration due to gravity is a constant.
Starting with the assumption of constant acceleration, we can write:
\[a = \frac{v_f  v_o}{t}\]
This can be rearranged to this form:
\[ta = v_f  v_o \longrightarrow ta + v_o = v_f \longrightarrow v_f = v_o + ta \]
We define the average velocity as:
\[v_{avg} = \frac{v+ v_o}{2}\]
We define displacement as:
\[x = x_o + v_{avg}t\]
We can now substitute \(v_{avg}\) with the previous equation.
\[x = x_o + \frac{v + v_ot}{2}\]
One more substitution for \(v = v_o+ at\) and we get:
\[x= x_o + v_ot + (½)at^2\]
We now have the four kinematic equations of motion.
\[v = v_o + at\]
\[x  x_o = (½)(v + v_o)t\]
\[x  x_o = v_ot + (½) at^2\]
\[v^2 = v_o^2 + 2a(x  x_o)\]
The first three equations relate displacement, velocity, and acceleration in terms of time, while the fourth equation does not contain the time.
Example 1
A train starts from rest and moves with constant acceleration. On first observation, the velocity is 20 m/s and 80 s later the velocity is 60 m/s. At 80 s, calculate the position, average velocity, and constant acceleration over the interval.
Calculate the acceleration:
\[a =\frac{v_fv_o}{t} = \frac{60\text{ m/s}  20\text{ m/s}}{80\text{ s}} = 0.50 \text{ m/s^2}\]
Calculate the distance traveled over this 80 s:
\[x = v_ot + (½)at^2 = (20\text{ m/s})(80\text{ s}) + (½)(0.50\text{ m/s^2})6400\text{ s^2}= 3200\text{ m}\]
The average velocity is:
\[v_{avg} = \frac{v_f+v_o}{2} = {20\text{ m/s}+60\text{ m/s}}{2} = 40\text{ m/s}\]
If the acceleration is constant, then the average velocity is the average of 20 m/s and 60 m/s, or 40 m/s, and at an average velocity of 40 m/s and 80 s, the distance traveled is 3200 m.
Example 2
Using the above problem, calculate the position of the train at 20 s.
\[x = v_ot + (½)at^2 = (20\text{ m/s})(20\text{ s}) + (½)(0.50\text{ m/s^2})(20\text{ s})^2 = 500 \text{ m}\]
Example 3
Using the above problem, find the time required for the train to reach 100m.
\[xx_o = v_ot+(½)at^2 \longrightarrow 100\text{ m} = (20\text{ m/s})t + (½)(0.50\text{ m/s^2})t^2\]
\[t = \frac{80{\pm} \sqrt{64004(1)(400)}}{2(1)} = \frac{80{\pm}89}{2} = 4.5, 85\]
The negative answer is not correct because we do not want negative time, so t=4.5s
Example 4
Using the above problem, find the velocity of the train at 120 m.
\[v^2 = v_o^2 + 2ax = (20\text{ m/s})^2 + 2(0.50\text{ m/s^2})120m = 400\text{ m^2/s^2} + 120\text{ m^2/s^2} = 520\text{ m^2/s^2} = 23\text{ m/s}\]
Example 5
Two vehicles are at position x=0 and t=0. Vehicle 1 is moving at constant velocity of \(30 \text{ m/s}\). Vehicle 2, starting from rest, has acceleration of \(10 \text{ m/s^2}\). Where do they pass?
Translated into algebra, this is asking what is the value of x when they pass? This can be determined by writing equations for the position of each vehicle and equates
\[ x_1 = v_10 t = 30 \text{ m/s} \text{ t} \]
and
\[x_2=(½)a_2t^2=(5\text{ m/s^2})t^2\]
Setting x1=x2 gives the time when they pass as \(30t=5t^2\) or \(5t(t6)=0\), so the vehicles pass at t=0 and t=6. Putting t=6 in either equation for x gives 180 m as the distance.
Example 6
In the above situation, when do the vehicles have the same velocity?
In algebra, this means to set the equations for velocity equal (v1=v2) and solve for the time. Remember that three of the four equations of motion are functions of time, so most questions are answered by first calculating the time for a certain condition to occur.
\[30\text{ m/s}=(10\text{ m/s^2})t = 3.0\text{ s}\]
Now that we know when, we can calculate where they have the same velocity. Use either equation for position and t=3.0s
\[x_1{t=3.0s}=(30\text{ m/s})(3.0s)=90\text{ m}\]
Example 7
For the situation in problem 5, what are the position, velocity, and acceleration of each vehicle when vehicle 2 has traveled twice the distance of vehicle 1?
The time when this occurs is when x2=2x1 so:
\[5t^2=60t\]
Or
\[5t(t12)=0\]
This gives the time of t=0 and t=12. The time t=0 is correct but t=12 is what we are looking for because we want to know something after some time has passed. At t=12,
\[x_1=(30\text{ m/s})(12\text{ s})=360\text{ m}\]
From the original statement of the problem,
\[v_1 0=30\text{ m/s}\]
and \(a_1=0\). Now solve for the remaining variables for the second vehicle by substitution.
\[x_2=(5\text{ m/s^2})(12\text{ s})=720\text{ m}\]
\[v_2=(10\text{ m/s^2})(12\text{ s})=120\text{ m/s}\]
From the original statement of the problem, \(a_2=10\text{ m/s}\).
Example 8
Two trains are traveling along a straight track, one behind the other. The first train is traveling at \(12\text{ m/s}\). The second train, approaching from the rear, is traveling at \(20\text{ m/s}\). When the second train is 200 m behind the first, the operator applies the brakes, producing a constant deceleration of \(0.20\text{ m/s^2}\). Will the trains collide, and if so, where and when?
First, diagram the situation. Write down the equations for each train and the information provided in the problem.
Train 1  Train 2 
\(x1=200m +(12\text{ m/s})t\)  \(x2=(20 \text{ m/s})t(0.10 \text{ m/s^2})t^2\) 
\(v1=12 \text{ m/s}\)  \(v2=20 \text{ m/s}(0.20 \text{ m/s^2})t\) 
\(a1=0\)  \(a2=0.20 \text{ m/s^2}\) 


Take t=0 when the brakes are applied and the first train is 200 m ahead of the second. This makes the position of the first train 200 m at t=0.
The question as to whether the trains collide means, is there a real time solution to the equation resulting from setting x1=x2?
\[200m+(12 \text{ m/s})t=(20 \text{ m/s})t(0.10 \text{ m/s^2})t^2\]
If there are no real solutions to the equation, then the trains do not collide. Drop the units, and write \(0.10t^28t+200=0\) which is solved by the quadratic formula. In a calculator, you can see there are no real solutions, so the trains never collide.
Example 9
Change the previous problem by giving the second train an initial velocity of 25 m/s. This will give a real time for the collision. Find the collision time.
Setting the expressions for x1 ands x2 equal and dropping the units produces
\[200+12t=25t0.10t^2\]
Or
\[0.10t^213t+200=0\]
This equals \(17.8, 112\)
Train 1  Train 2 
\(x_1=200 \text{ m}+(12 \text{ m/s})t\)  \(x_2=(25 \text{ m/s})t(0.10 \text{ m/s^2})t^2\) 
\(v_1=12 \text{ m/s}\)  \(v_2=25 \text{ m/s}(0.20 \text{ m/s^2})t\) 
\(a_1=0\)  \(a_2=0.20 \text{ m/s^2}\) 
The two times correspond to when x1=x2. The earliest time is the first coincidence and the end of the physical problem. The position at this time can be obtained from either expression for x.
\[x_1=200m+(12 \text{ m/s})17.8\text{ s}=414 \text{ m}\]
Verify this distance by using x_2. The velocity of the second train at collision is
\[v_2=25 \text{ m/s}(0.20 \text{ m/s^2})17.8\text{ s}=21.4 \text{ m/s}\]
The relative velocity between the two trains is \(v=1.4 \text{ m/s}\). The two times are the result of the quadratic in t. The two solutions occur when the curves cross. The equation for \(x_1=200+12t\) is a straight line of slope 12 starting at 200. The equation for \(x_2=25t0.10t^2\) is a parabola that opens down. While the mathematics produces two times, the reality of the problem dictates the earlier time as the one for the collision.