# Physics with Calculus/Mechanics/Motion in One Dimension

## Motion in One Dimension

As physics, at its essence, is the study of motion, we will begin our book with the study of motion, in one dimension.

To simplify our discussion, we will want to create a model of the real objects we will be discussing. Thus, we create the concept of a "particle". A particle, simply, is a very small object, one whose length, width, and height are negligible. With this concept of a particle, we can unambiguously define the distance from it to the origin, without having to worry about from what point we measure the object.

### Position

Consider a particle restricted to movement along a single line. Its position can be described with a single number, x. This number can be thought of as the particle's distance from the origin, or the magnitude of the vector pointing from the origin to the particle. The actual displacement of the particle is actually a vector with one entry (because we are working in one dimensional space), but because the direction is known it can be specified unambiguously by the magnitude alone.

The value x is measured in meters, and in order to interpret it, a particle and a reference frame, or coordinate system, must be implied. If the particle does not remain in the same place at different times, then the value of x will change in time, and thus we can define a function x(t) which returns the position of the particle at a point in time t.

For example, say we have a particle with position function x(t)= 2t. Its position at t = 3 is x(3) = 2(3) = 6 m along the x-axis.

For a particle with position function x(t)= t2 + 5, its position at t = 7 s is x(7) = 72 + 5 = 54 m along the x-axis.

### Velocity

Consider the graph of a position function. Under any known physical conditions, this function must be continuous — if it were discontinuous, that would allow the possibility that objects could spontaneously jump from place to place. Furthermore, as we shall see at the end of this section, the position function is also twice differentiable. Under these conditions, we are able to see that the function has a well defined rate of change everywhere.

Velocity, in one dimension, is the measure of the rate of change position per unit time, and is measured in meters per second. The simplest way to measure velocity is to measure the time position of a particle at two different times, and divide the change in position (displacement) by the change in time. As a formula:

${\bar {v}}={\frac {\Delta x}{\Delta t}}={\frac {x_{f}-x_{i}}{t_{f}-t_{i}}}$

where xf and xi are the final and initial distances respectively and tf and ti are the final and initial times.

Notice that since the numerator has units of length and the denominator has units of time, the quotient has units of length divided by time. In SI, the units of velocity are the derived units of meters per second. Common non-SI units of velocity include miles per hour and knots.

Example 1: A particle starting at the origin has moved 27 m in 9 s. Find ${\bar {v}}$ .

${\bar {v}}={\frac {x_{f}-x_{i}}{t_{f}-t_{i}}}={\frac {27\ \mathrm {m} -0\ \mathrm {m} }{9\ \mathrm {s} -0\ \mathrm {s} }}=3\ \mathrm {m} /\mathrm {s}$

Example 2: A particle starting 23 m from the origin has moved to 43 m in 5 s. Find ${\bar {v}}$ .

${\bar {v}}={\frac {x_{f}-x_{i}}{t_{f}-t_{i}}}={\frac {43\ \mathrm {m} -23\ \mathrm {m} }{5\ \mathrm {s} -0\ \mathrm {s} }}={\frac {20\ \mathrm {m} }{5\ \mathrm {s} }}=4\ \mathrm {m} /\mathrm {s}$

There is a limit, however, to what average velocity can tell us. For example, it's entirely possible that the speed of a particle varies with respect to t, and we want to determine the velocity of the particle at a certain instant in time. To do this, we take the average velocity over a smaller and smaller time interval, that is,

$\ v(t)=\lim _{\Delta t\to 0}{\frac {\Delta x}{\Delta t}}$

Expanding this, we get

$v(t)=\lim _{\Delta t\to 0}{\frac {x(t_{f})-x(t_{i})}{t_{f}-t_{i}}}$

And rewriting $t_{i}$  and $t_{f}$  as t and t+$\Delta t$ , we get

$v(t)=\lim _{\Delta t\to 0}{\frac {x(t+\Delta t)-x(t)}{\Delta t}}$

Which is the definition of the derivative. Therefore,

$\ v(t)={\frac {dx}{dt}}$

Example 1: A particle's motion is described by the equation $\ x(t)=3t^{2}+5t+2$ . Find a) the particle's velocity function, and b) its instantaneous velocity at t = 4 s.

a) $v(t)={\frac {dx}{dt}}={\frac {d}{dt}}(3t^{2}+5t+2)=6t+5$

b) $\ v(4)=6(4\ \mathrm {s} )+5=29\ \mathrm {m} /\mathrm {s}$

### Acceleration

Just as the time derivative of a particle's position function has a significant meaning, that is, its velocity, so too does the time derivative of the velocity function. In this case, it is known as the acceleration of the particle. You may have heard of acceleration previously, such as when referring to cars or other vehicles. Acceleration is measured in meters per second per second, or m/s2, read 'meters per second squared.' Some old texts use the equivalent phrase 'meters per second per second.'

Average acceleration, or ${\bar {a}}_{x}$  is similar to average velocity in that it is independent of the path taken by the particle; it is simply the change in velocity divided by the change in time:

${\bar {a}}={\frac {v_{xf}-v_{xi}}{t_{f}-t_{i}}}={\frac {\Delta v_{x}}{\Delta t}}$

And again, through a similar derivation, we arrive at the definition of instantaneous acceleration:

$a(t)=\lim _{t\to 0}{\frac {\Delta v}{\Delta t}}={\frac {dv}{dt}}$

Since the velocity is a derivative of the original position function, we can also say that the instantaneous acceleration is the second derivative of the position function, which would be written as follows:

$a(t)={\frac {d^{2}x}{dt^{2}}}$

### Exercises

1. Given a particle moving in one dimension with position function x(t) = 5t3 – 8t2 + 20:
1. Determine the velocity v as a function of time t.
2. Determine the acceleration a as a function of time t.
3. At what time(s) will the particle come to a momentary stop?
4. At what time(s) will the particle's acceleration be zero?
2. A particle in simple harmonic motion has a position defined by the function x(t) = (80 cm) sin [(20 s-1)t].
1. Determine the velocity v as a function of time t.
2. Determine the acceleration a as a function of time t.
3. What is the period of the particle's motion? (That is, how long does it take for the particle to return to the position and velocity at which it started?)
3. Excluding the effects of air resistance, an apple falls from a tree with a constant acceleration of 9.8 m/s2.
1. What is the apple's velocity v(t)?
2. What is the apple's velocity after one second of freefall?
3. What is the position x(t)?
4. What is the position after one second of freefall? Two seconds? Three seconds?

## Motion in Two or Three Dimensions

Describing the motion of a particle in two or more dimensions can be more complex than in one dimension, because, instead of having only two directions in which to move (back and forth along the x-axis) there are an infinite number of paths a particle can take. In order to properly describe the motion of a particle, we must use vectors.

### Position

When dealing with motion in one dimension, we could describe an object's position with a single variable, its location along the x-axis. Now that we have two or more axes, we need one variable per axis to describe a particle's location. We can use a position vector, denoted r, to describe the location of this particle. In the two-dimensional case, the position vector is drawn from the origin to the location of the particle in the Cartesian plane and has two coordinates, the x-position of the particle, and the y-position of the particle.

Example 1: The position of a particle at (4, 5) on a Cartesian plane can be described as 4 units along the x-axis, and 5 units along the y-axis, or as the position vector r = <4, 5> = 4i + 5j, where i and j are the unit vectors in the x and y direction, respectively.

From this example, we can see that a position vector takes the form <x, y>, where x is the distance moved along the x-axis and y is the distance moved along the y-axis.

Example 2: A carbon atom sits 7.0 nm east, 0.5 nm north, and 3.0 nm below the tip of a scanning tunneling microscope. Its position vector r = <7.0 nm, 0.5 nm, –3.0 nm>. Taking i to be east, j to be north, and k to be up, r = (7.0 nm)i + (0.5 nm)j + (–3.0 nm)k.

### Displacement

Recall that the displacement of a particle is the difference between its initial and final positions. In two dimensional motion, we denote the displacement vector as Δr. The displacement vector is defined as:

$\Delta \mathbf {r} =\mathbf {r} _{f}-\mathbf {r} _{i}$

The direction of the displacement vector is the straight line between the starting point and the termination point of a particle's path.

### Velocity

As with motion in one dimension, we still retain the concepts of average and instantaneous velocity; the only difference is that we now must treat the change in position as a change in the position vector. Thus, the average velocity in two dimensions is

${\bar {\mathbf {v} }}={\frac {\Delta \mathbf {r} }{\Delta t}}$

Since, like displacement, ${\bar {\mathbf {v} }}$  is independent of path, the direction of ${\bar {\mathbf {v} }}$  is along $\mathbf {r}$ .

Like its one-dimensional analog, the instantaneous velocity v is calculated by finding the time derivative of the vector displacement r:

$\mathbf {v} ={\frac {d\mathbf {r} }{dt}}$  COUT <<SAFA;

### Acceleration

Just as with the one-dimensional case, the average and instantaneous accelerations can be calculated using derivatives and limits.

The average acceleration is given by: ${\bar {\mathbf {a} }}={\frac {\mathbf {v} _{f}-\mathbf {v} _{i}}{t_{f}-t_{i}}}={\frac {\Delta \mathbf {v} }{\Delta t}}$

The instantaneous acceleration is given by: $\mathbf {a} ={\frac {d\mathbf {v} }{dt}}={\frac {d^{2}\mathbf {r} }{dt^{2}}}$ .