Section 1-2 - Kinematics in One Dimension

Displacement and Distance

We'll need first of all to be able to define the location of an object.  In one dimension, we can consider the 'numberline' axis discussed in the last section and use the variable 'x' to label the position relative to the origin in metres, so that the statements 'x = +3 m' and 'x = -7.46 m' mean that the object is 3 m from the origin in the positive direction (not necessarily to the right of it, though!) and 7.46 m from the origin in the negative direction, respectively.
We also need to be able to describe the change in location of an object.  We define the displacementDx of an object which started at x_i and ended up at x_f as
Dx = x_f - x_i.
Note that, according to this definition, the displacement depends only on where the object started and ended, not on the path taken.
Examples:  Suppose that an object starts out at x_i = 3 and ends at x_f = 5, and makes that trip smoothly and without reversing direction.  What is the displacement?  Now, suppose instead that the object travels from x = 3 to x = 15, then back to x = - 8, then on to x = 5.  What is the displacement in that case?

Distance (s) is the term we use for the length of the path taken.  So long as the direction of motion doesn't change, the distance is the same as the magnitude of the displacement.
Examples:  Suppose that an object starts out at x_i = 3 and ends at x_f = 5, and makes that trip smoothly and without reversing direction.  What is the distance?  Now, suppose instead that the object travels from x = 3 to x = 15, then back to x = - 8, then on to x = 5.  What is the distance in that case?

Velocity and Speed

Average speed is defined as the distance traveled per unit time, or speed = s/Dt.  Find the average speed in these examples.
Suppose that an object starts out at x_i = 3 and ends at x_f = 5, and makes that trip smoothly and without reversing direction in 3 seconds.
Suppose instead that the object travels from x = 3 to x = 15, then back to x = - 8, then on to x = 5, all in 3 seconds.

Try another example:
Jimmy walks across the room (10 m) in 10 seconds, and runs back in 5 seconds.
What is his displacement?

The average velocity discussed above is considered over an interval of time.  How can we find the instantaneous velocity, the velocity at an instant of time?  Consider the following graph, which shows the position of an object as a function of time, x(t):

How would the average velocity between t_o and t_f be represented on this graph?
v_ave = Dx/Dt = [x_f - x_i]/[t_f - t_i] = rise over run = slope of the line connecting the two points.

How to find the velocity at time t_i?  Let's decrease the interval:

As the interval becomes smaller, the average velocity approaches the instantaneous velocity, or graphically, the slope representing the average velocity approaches the slope of the line tangent to the x(t) curve at the point at which we wish to know v(t).  Mathematically, we write this as
v_{instantaneous} = limit _Dt->0Dx/Dt.
 

We can do the same with the instantaneous speed:
inst speed = limit _Dt->0 s/Dt.
We then see that, for infinitely short time intervals, an object doesn't have time to reverse direction, and so the instantaneous speed is the same as the magnitude of the instantaneous velocity. 

Acceleration and Jerk

We can continue the process indefinitely.  For example, the average jerk is defined as
j_ave = Da/Dt,
and the instantaneous jerk is
j_inst = limit _Dt->0Da/Dt,
and so on with the kick and then the lurch:
k_ave = Dj/Dt,     the instantaneous kick is k_inst = limit _Dt->0Dj/Dt,
l_ave = Dk/Dt,     the instantaneous lurch is l_inst = limit _Dt->0Dk/Dt.

The acceleration, jerk, kick, and lurch are all vector quantities.

Kinematic Equations

• that the acceleration is constant, at least over some interval in which we are interested.
• that the problem starts at t_i = 0, so that we can just drop that term.
• that all final quantities (t_f, x_f, v_f) can be  replaced with the more general corresponding variables (t, x, v).

Start with the definition of the acceleration (since the acceleration is constant, that value is also the average value):
a = [v - v_o]/[t - t_i] = [v - v_i]/t,
which re-arranges to
v = v_i + at.
Strictly speaking, this is true so long as a is the average acceleration.

Before we continue, we need to develop a new relationship.  Consider the v(t) graph below:

The curve is a straight line, because the acceleration is constant and is represented by the slope of the curve; it may well have been a negative slope, or even a zero slope, instead of the positive slope pictured here.  We need to average the infinitely many values the velocity has in the interval t_i to t_f.

We can do it without calculus if we're a little clever:  First, average just the two endpoints to get [v_i + v_f]/2. 
Now, average the points just above v_o by an amount e and just below v_f by the same amount e to get 
[(v_i + e) + (v_f - e)]/2 = [v_i + v_f]/2. 
So e can have any value and result in the same average value for any given pair of symetrically placed points.  Since the overall average is the average of the pairs' averages, it should be clear that the overall average will also be 
[v_i + v_f]/2. 
So, for periods of constant acceleration, v_ave = [v_i + v]/2.

Let's start again with the definition of average velocity, 
v_ave = (x - xi)/t, 
which re-arranges to 
x = x_i + v_avet. 
Now, substitute the new expression for v_ave to get 
x = x_o + ([v_i + v]/2)t. 
Previously, we showed that 
v = v_i + at. 
Let's substitute this expression into the one above it 
x = x_i + ([v_i + v_i + at]/2)t 
and simplify: 
x = x_i + v_it + ¹/₂at². 

Let's start again with the definition of the average acceleration:
a = [v - v_i]/t,
which we can re-arrange to be
t = [v - v_i]/a.
Remember that
v_ave = [v + v_i]/2.
Now, substitute each of these into
x = x_i + v_avet.
x = x_i + ([v + vi]/2)([v - v_i]/a)
Now, simplify to get
x = x_i + [v² - v_i²]/2a.
Re-arrange to obtain
v² = v_i² + 2a(x -  xi).

Now, we have four kinematic equations that are valid in the special case of constant acceleration:

v = vi + at vave = [vi + v]/2 x = xo + vit + 1/2at2 v2 = vi2 + 2a(x -  xi)

Various combinations and perturbations of these should allow for solving most problems.  Here, however, is a warning: do not rely on the equations by themselves to solve problems.  The equations are in a sense tools, but it still requires the brain to direct their use.

Example:
A distracted driver traveling at 15 m/s notices a stop sign when he is 10 m from the stop line.  If the car decelerates at 6 m/s², how quickly is the car moving as it passes the stop line?

Let's write down the quantities which we know either implicitly or explicitly, as well as what we want to figure out: 

Let positive x be in the direction the car is moving and the origin be where the driver first applies the brakes.
x_i = 0 m
x_f   = 10 m
v_i = 15 m/s
v_f  = ?
a = - 6 m/s² (a deceleration of 6 m/s² is an acceleration of -6 m/s², since a velocity becoming less positive is the same as one becoming more negative).

Since the kinematic equations are really all the same relationships presented in slightly different forms, we can look for one which contains all of the quantities above.  Sometimes this works, sometimes not; in this case we're lucky:
v_f² = v_o² + 2a(x -  x_o),
and in fact, not much algebraic manipulation is necessary:
v_f² = v_o² + 2a(x -  x_o) =  15² + 2(-6)(10) = 105
v_f = 105^1/2 = 10.2 m/s
Note that we took the positive root of 105.  Strictly speaking, the math will only give us the final speed in this case; we need to use our brains to determine the sign (and hence the direction) of the final velocity.

Let's ask a slightly different problem.  How far into teh intersection does the car actually stop?  The values we use for this are then
x_o = 0 m
x_f   = ?
v_o = 15 m/s
v_f  = 0 m/s
a = - 6 m/s²

Since we have the same pieces of data, as in the last example, let's try the same equation:
v_f² = v_o² + 2a(x -  x_o)
x = x_o + (v_f² - v_o² )/2a = 0 + (0² - 15² )/2(-6) =  18.75 m.
However, this is not the answer!  We were asked how far into the intersection te car stops, but thsi 18.75 m is measured from where the brakes were first applied.  The correct answer is 8.75 m into the intersection.

Mastery Question
A train moving at 15 m/s passes the origin (x_i = 0) at t_i = 0.  At that instant, the engineer hits the brakes, giving the train an acceleration of - 0.5 m/s², so that it comes to a stop.  Where is the train after 40 seconds?

Click here for the solution.

What should we do when the acceleration is not constant?  So long as it is constant over intervals and changes abruptly, we can treat each individual interval as above, using the final values of the quantities in one interval as the initial values for the next interval.  See, for example, Problem 2-7.

Acceleration due to Gravity

The results of an experiment by the Fall 2003 PHY542 class are shown below.  After dropping a mass from rest (v_i = 0) through vertical displacement h and measuring the travel time, the data were plotted as h vs t².  If the kinematic relationships are true, the slope of this curve represents half of the acceleration due to gravity, a_g.  A value of 9.8126 m/s² is within about 0.14% of the accepted value in Towson.

In most problems, you may feel free to use 10 m/s² as the value for a_g.

Example:
Let's drop a water balloon on someone's head from the top of a 20m tall building.  How quickly will the balloon be moving at the bottom and how long will it take to arrive there?

Let up be positive and the origin be the top of the building.  Then,
x_i = 0 m
x_f   = -20 m
v_i = 05 m/s
v_f  = ?
a = -10 m/s²
t = ?

Again, let's check to see if there is a single kinematic equation that will give us the answer, and there is:
v_f² = v_i² + 2a(x -  x_i),
v_f² = v_i² + 2a(x -  x_i) =  0² + 2(-10)(-20) = 400
v_f = 400^1/2 = -20 m/s
We must take the negative root here, because the balloon is moving in the negative direction at the end of the problem.  The equation doesn't know to do that; we need to keep an eye out.
As for the time, the shortest path is to make use of the final velocity above and use
v = v_i + at
t = (v - v_i)/a = (-20 - 0)/(-10) = 2 seconds.

Or we might have used
x = x_i + v_it + ¹/₂at²
-20 = 0 + 0t + ¹/₂(-10)t²
t² = (-2)/(-5) = 4     
t = +2 seconds 
We take the positive root because the balloon hits the ground after it's dropped.

Example with solution:
A ball is thrown from the street such that it rises past a 25m high window ledge at 12 m/s.  Find a) the velocity with which it was launched, b) the maximum altitude above the street it reaches, c) how long ago it was thrown, and d) the time until it returns to the ground.  Click here for solution.

A Different Graphical Interpretation

Of course, in this example, we are always underestimating the displacement, because we're multiplying Dt by the (lower than average) initial velocity of each interval.  So, we want to make as many intervals as possible, each over as small a time interval as possible, to reduce this error.  As we let the number of intervals go towards infinity, we can see that the little triangles atop each rectangle get smaller and smaller, and that the total rectangle area we are counting tends toward the total area under the line.  So we see that the total displacement will be represented on such a graph as the area under the curve. 

Since the original shape was a trapezoid, the area of which is 
¹/₂[h₁ + h₂]b, 
the displacement will be 
¹/₂[v_i + v_f] Dt. 
Can we verify this from what we already knew?  The displacement (by re-arrangement of a definition) is Dx = v_ave Dt.  But we calculated v_ave for constant acceleration mathematically: 
v_ave = [v_f + v_i]/2. 
So, 
Dx = v_ave Dt = ([v_f + v_i]/2) Dt 
as expected. 

We can generalize this result for any shape of curve: 

Also, we can use exactly the same argument to assert that the area under the acceleration vs time curve is the change in velocity.  That bears repeating: we can't get the velocity from the curve, only thechange in velocity, in the same way that we got the displacement, the change in position, for the v vs t curve, not the object's position itself.

Here is a quick example:

Each of the two curves shown will generate the same acceleration curve, since the slopes of the two are the same for each value of time, t.  So, given a particular acceleration curve, it would be impossible for one to determine which of an infinite number of velocity curves it was derived from.

Let's look at another situation:

In this case, the velocity starts out positive, but there is a negative acceleration (slope of the line).  Eventually, the velocity becomes zero and the object comes momentarily to rest, having traveled through a displacement represented by the area under the curve (the red area).  As time progresses, we see that the velocity becomes negative, the object reverses direction, and we would expect that it may well arrive back at its starting point, for a total displacement of zero.  How does this play out on the graph?  Since displacement is basically velocity times time interval, negative velocities result in negative inceremental displacements which are represented by negative area (blue).  So, in this example, when the positive (red) area above the axis equals the negative (blue) area under the axis, the total displacement will be zero and the object will have returned to its starting point.

Consider a mass oscillating on a spring:

Once again, when the area 'under' the curve adds to zero, the object has returned to its starting point.

Discussion of Misconceptions and the Act of Misconceiving

Ball dropped from a rising helicopter (see Problem 2-8):
Many students believe that the ball begins to descend immediately upon its release from the rising helicopter.  This is not so.  Try lifting a pen upward with your hand palm down, releasing it as it passes some point on the wall.  If that notion is correct, the pen will never appear above that spot on the wall, but you will see that it does indeed continue to rise.  For some reason, it's easier to believe when you push the pen with palm upward, so be sure to hold palm downward to more closely simulate the scenario in the problem.  Graphically, we see

We see at first the constant velocity experienced while the ball is still connected to the helicopter.  At the release time, the acceleration becomes -9.8 m/s², as shown by the negative slope of the line.  A short time later, the velocity is still positive, although less than it was, but a positive velocity still means that the object is rising, and it will continue to do so until the velocity reaches zero at the highest point in the trajectory.

Speaking of that highest point, what is the acceleration at that point?  It is common to assume that it is zero, but that confuses that velocity with the acceleration.  We can look at the graph above and see that the slope of the v(t) graph when v = 0 is still -9.8 m/s².  Or, think opf the velocity just before the peak (positive) and just after the peak ( negative); the acceleration measures the change in velcoity, which became more negative in that time interval.  Or think of it this way, since acceleration is related to the change in velocity, if a_g were zero at the peak, what would the object do?  No acceleration implies no change in velocity, so the object would just hang in space, an event counter to our experience.