# Mechanics 2: Motion in One Dimension

## Motion in One Dimension

Last chapter, we started with *position* (\(x\) or \(y\)), then used a derivative to get *velocity* (\(v\)), and then a second derivative took us to *acceleration* (\(a\)). But what if we wanted to go the other way, i.e., start with acceleration and work to position? We can do that, but we need to learn another math “recipe.” One name for it is an *antiderivative.*

So, as you may have guessed, an antiderivative is the inverse operation of a derivative. The recipe is: increase the exponent, then divide by the new power. Example:

$${\rm antiderivative}\ (t^3) = \frac{t^4}4$$

Does this make sense? We can check by taking the derivative of \((t^4/4)\):

$$ \frac{d}{dt} \bigg( \frac{t^4}4 \bigg) = \frac{4t^3}4 = t^3 $$

giving us back what we started with, \(t^3.\)

There’s one more part to this “recipe.” Remember that the derivative of a constant is zero, so the antiderivative of 0 is a constant. To account for this, we need to add an arbitrary constant. Usually, it’s called, \(C,\) or if we need multiple constants, \(C_1, \ C_2,\) etc. So, the antiderivative above should be

$${\rm antiderivative}\ (t^3) = \frac{t^4}4 + C$$

And, if we take the derivative of this, we get back \(t^3,\) same as above.

OK, let’s give this “recipe” a name, describe what it represents, and learn its “official” notation. Remember that a derivative is the slope of the curve. The antiderivative, most commonly called the *integral,* is the area under the curve (we’ll talk more about this later). Since the integral of a function is the area under the curve, it’s like summing up a bunch of little areas that fit under the curve, to add up the area under the curve. Since the integral represents a sum, its symbol is an “elongated S”. This equation would be read as, “the integral of \(t^3\) with respect to \(t\)” and is written like so:

$$\int t^3\,dt = \frac{t^4}4 + C$$

Just as for derivatives, the first few powers of \(t\) and their integrals are shown; note that below, \(a\) is a constant:

\(f(t)\) | \(\int f(t)\,dt\) |

\(0\) | \(C\) |

\(a\) | \(at + C \) |

\(t^1\) | \(\frac{t^2}2 + C\) |

\(t^2\) | \(\frac{t^3}3 + C\) |

\(t^3\) | \(\frac{t^4}4 + C\) |

and so on. This “recipe” only works for polynomials — other functions like sines and cosines we’ll get to later.

So how do we use this? As you might be able to guess, this time the progression is like this

$${\rm position} \stackrel{\rm integral}{\longleftarrow} \ {\rm velocity} \stackrel{\rm integral}{\longleftarrow} {\rm acceleration}$$

We’ll start by calculating the equations of motion for something that has constant acceleration — that means that the acceleration doesn’t change with time (like gravity). That means acceleration is a fixed value for all time, and can be zero. We’ll call it \(a\) for now. So, we can work from acceleration to position by integrating over time, like this:

$$ a(t) = a $$

$$ v(t) = \int a\,dt = at + C_1$$

Since \(a\) is a constant, its integral is \(at,\) then we have to add the arbitrary constant, \(C_1.\) We use a subscript one here, because we’ll need a second constant when we calculate a second integral, in the next step.

Now, when we integrate (calculate the integral) a second time, we get:

$$ x(t) = \int (at + C_1)\,dt = \frac12 at^2 + C_1t + C_2 $$

Remember, the integral of \(at\) is \(\frac12 at^2\) and the integral of a constant (\(C_1\) here) is just the constant times \(t.\) And again, we have to add an arbitrary constant — call it \(C_2.\)

Next, we need to figure out what these constants (\(C_1\) & \(C_2\)) should be. There are many ways of handling these arbitrary constants.

What we need are called “constraints” or particular conditions. There are two kinds: initial conditions (at \(t=0\)) or boundary conditions (what happens at the edges in \(x\) or \(y\)). For our example, we’ll use initial conditions. So, we’ll call the initial velocity, \(v(t=0) = v_0.\) This is read, “v-naught” (the British word for zero). Then, the initial position, \(x(t=0) = x_0\) (read as “x-naught”). Next, we’ll see how to use these conditions.

Take the velocity equation:

$$v(t) = at + C_1$$

Our initial condition tells that at \(t=0,\) \(v = v_0,\) giving

$$v(t=0) = a \cdot 0 + C_1 = v_0$$

For the first term, \(a \cdot 0 = 0,\) so \(C_1 = v_0.\) Now let’s do a similar thing with the position equation:

$$x(t) = \frac12 at^2 + C_1t + C_2 $$

This time we use the other initial condition of \(x=x_0\) at \(t=0,\) and we get

$$x(t=0) = \frac12 a \cdot 0 + C_1 \cdot 0 + C_2 = x_0$$

The first two terms are 0, so \(C_2 = x_0.\) Now collect all this together, and we get the equations of motion for constant acceleration (i.e., \(a\) is constant):

\begin{align}

a(t) &= a\\

v(t) &= at + v_0\\

x(t) &= \frac12 at^2 + v_0t + x_0

\end{align}

### Examples

Let’s take a look at an example of how to use these equations of motion. Say you were on top of a house, repairing something, and a tool slipped out of your hand. How long would it take to hit the ground, and how fast would it be moving? Let’s say that the peak of your house is 10 m high. So, the initial position would be \(y_0 = 10\) m. Since you “let go” of the tool, its initial velocity, \(v_0 = 0.\) So, the equations of motion in the vertical (\(y\)) direction are

$$a(t) = -g$$

constant acceleration downward (shown by the negative sign) only due to gravity. If we integrate that, we get

$$v(t) = -gt + v_0 = -gt$$

Since \(v_0 = 0,\) the constant term goes away. Finally, integrate again to get

$$y(t) = -\frac12 gt^2 + y_0 = -\frac12 gt^2 + 10$$

Now what? To answer the first question about how long — the ground is at \(y=0,\) so set the last equation \(=0\) and solve for \(t.\)

$$0 = -\frac12gt^2 + 10$$

or

$$\frac12gt^2 = 10$$

and

$$t^2 = \frac{10}{\frac12g} = \frac{2 \times 10}g$$

The acceleration due to gravity, in MKS units, is \(g=9.8\) m/s\(^2,\) which gives

$$t^2 = \frac{2 \times 10}{9.8} = 2.04$$

and then taking a square root, \(t = 1.43 \)s.

For the next question, how fast is it moving when it hits, use the time we just found, and plug that into the velocity equation, like so

$$v(1.43) = -gt = -9.8 \times 1.43 = -14\,{\rm m/s}$$

This tells us that when it hits, it is moving downward at 14 m/s.

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