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2. One Dimensional Kinematics

The study of motion of objects is called kinematics. If an object moves along a straight line, the motion is said to be in one dimensions. We will start our study of kinematics by considering simple 1D motion. However, in our everyday experiences we see common examples of motions in two and three dimensions. For examples, a car moving on a flat road which may be curved or an object thrown in the air at an angle relative to the ground (projectile motion) are motions in two dimensions; we consider two-dimensional motion in the next section.

2.1 Time and position

We will use \(t\) to denote time. To describe a one-dimensional motion, only one extra variable is needed for the position. We will use \(x\) to denote the position for a one-dimensional motion.

In the widget above, you can see a rectangular object in a simple 1D motion from left to right.

Each grid on the screen represents one unit of distance -- in this case one grid represents \(1\ cm\). In this particular example, the motion of the box is such that it moves exactly one unit in one second (unless you have changed the speed of the box using the slider).

2.2 Displacement

An object in one dimensional motion moves from an initial location \(x_i\) to a final location \(x_f\) in a certain time interval \(\Delta t = t_f - t_i\). Then, the difference of the final position and the initial position is known as the displacement: \(\Delta x = x_f - x_i\).

Examples

If \(x_i=0\ m\) and \(x_f = 1\ m\), then the displacement is \(\Delta x = x_f - x_i = 1\ m\).

However, if the particle is initially at \(x_i=1\ m\) and travels to the left to \(x_f = 0\ m\), then the displacement is \(\Delta x = x_f - x_i = -1\ m\).

In both examples above, the distance traveled is \(|\Delta x| = 1\ m\). The negative sign, in the second example, signifies the direction (left) of the displacement. Displacement is a vector quantity whereas distance is a scalar quantity.

2.3 Speed

Use the slider in the figure above to change the units of distance \(x\) moved by the moving object in one second. If you take the ratio of the displacement \(\Delta x\) in an interval of time \(\Delta t\), the quantity is called average speed. At a particular time \(t\), if you find the speed by taking \(\Delta t \rightarrow 0\), the quantity is called instantaneous speed.

Mathematically

\[ v_{\rm avg} \equiv \bar{v} = \frac{\Delta x}{\Delta t} \]
\[ v = \lim_{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t} = \frac{d x}{d t} \]
\[ \frac{d (c\ t^n)}{dt} = c\ n\ t^{n-1} \]

where \(c\) is a constant and \(n\) is an integer.

The SI unit for speed is \(m/s\).

Examples

Suppose a box moves from \(x=0\) to \(x=50\ cm\) in \(5\ s\), rests at \(x=50\ cm\) for \(2\ s\), and then moves from \(x=50\ cm\) to \(x=1\ m\) in \(1\ s\). What is the average speed?

Solution: From start to finish, \(\Delta x = 50\ cm + 0 \ cm + 50 \ cm = 100 \ cm\).

The time taken is \(\Delta t = 5\ s + 2\ s + 1\ s = 8\ s\).

Therefore, the average speed of the box is

\[ \bar{v} = \frac{\Delta x}{\Delta t} = \frac{100\ cm}{8\ s} = 12.5\ cm/s = 0.125\ m /s \]

Suppose that the position of a box is described by the following function of time:

\[ x(t) = 0.5\ m + (0.8\ m/s)\ t \]

The instantaneous speed at time \(t\) can be found by taking the derivative as follows:

\[ v(t) = \frac{d x(t)}{d t} = 0.8\ m/s \]

That is at each point of motion of the box the speed is \(0.8\ m/s\).

Suppose that the position of a box is described by the following function of time:

\[ x(t) = 0.5\ m + (5.8\ m/s)\ t - (4.9\ m/s^2)\ t^2 \]

The instantaneous speed at time \(t\) can be found by taking the derivative as follows:

\[ v(t) = \frac{d x(t)}{d t} = 5.8\ m/s - (9.8\ m/s^2)\ t \]

which is a function of \(t\) as well. That is at each point of motion of the box, the speed is changing. The speed at time \(t\) can be found by plugging in the value of \(t\) in the function \(v(t)\) calculated above. For examples:

  • at \(t=0\ s\), \(v(0) = 5.8\ m/s\)
  • at \(t=1\ s\), \(v(1\ s) = (5.8\ m/s)-(9.8\ m/s^2)(1\ s) = -4.0\ m/s\)

Notice that the expression for speed looks like a slope -- rise over run. Therefore, when you plot position \(x\) of a particle as a function of time \(t\), the slope gives the speed of the particle at time \(t\). For a constant speed case, the \(x\) vs \(t\) graph is linear.

See Example 3.2 in OpenStax.

2.4 Velocity

Velocity is a vector quantity i.e. it has both magnitude and direction. The magnitude of a velocity is the speed of the object, and the direction is the direction in which the object is moving.

In a 1D horizontal motion along the \(x-\)axis, only two directions are possible: to the left and to the right. These can be denoted by positive (to the right) and negative (to the left) values of speed. We will learn more about velocity vectors in the coming chapters.

2.5 Acceleration

Moving objects can experience a change in velocity. The rate of change of velocity is called acceleration. Acceleration is also a vector quantity.

Mathematically

\[ a_{\rm avg} = \frac{ \Delta v }{\Delta t} \]
\[ {a} = \lim_{\Delta t \rightarrow 0 } \frac{\Delta {v}}{\Delta t} = \frac{d {v}}{d t} \]

Examples

Suppose an object is moving to the right at \(t=0\) with a speed of \(10\ m/s\). It hits a wall and rebounds back to the left. The speed of the object is found to be \(5\ m/s\) when measured at \(t=3\ s\) as it is moving to the left. What is the average acceleration of the object in this interval?

Given quantities are: \(v_i = 10\ m/s\), \(v_f = -5\ m/s\), and \(\Delta t = 3\ s\).

\[ \Delta v = v_f - v_i = (-5\ m/s) - (10\ m/s) = (-15\ m/s) \]
\[ {a}_{\rm avg} = \frac{\Delta v}{\Delta t} = \frac{(-15\ m/s) }{3\ s} = (-5\ m/s^2)\ \]

Notice that the expression for acceleration also looks like a slope -- rise over run. Therefore, when you plot velocity \(v\) of a particle as a function of time \(t\), the slope gives the acceleration of the particle at time \(t\). For a constant acceleration case, the \(v\) vs \(t\) graph is linear.

See Figure 3.15(a,b) and the discussion around them in OpenStax textbook.

The area under the graph of a \(v\) vs \(t\) plot gives the displacement \(\Delta x\) in the given time interval.

2.6 Equations of motion

The following equations relate various quantities in 1D motion of an object undergoing a constant acceleration \(a\). The initial time is set to be \(t_0 = 0\), the initial speed is \(v(t=0) = v_0\) and the initial location is \(x(t=0)=x_0\).

List of equations and their proofs

\[ \begin{align} x &= x_0 + \bar{v} t \\ v &= v_0 + a t \\ \bar{v} &= \frac{v + v_0}{2} \\ x &= x_0 + v_0 t + \frac{1}{2} a t^2 \\ v^2 &= v_0^2 + 2 a \ \Delta x \end{align}\]
\[ v_{\rm avg} = \bar{v} = \frac{\Delta x}{\Delta t} \]
  • Also, \(\Delta t = t - t_0 = t - 0 = t\), and
  • \(\Delta x = x - x_0\), so that
\[ \bar{v} = \frac{x-x_0}{t} \implies x = x_0 + \bar{v} t \]
\[ a_{\rm avg} = \frac{\Delta v}{\Delta t} \]
  • For a constant acceleration motion \(a_{\rm avg} = a\).
  • Also, \(\Delta t = t - t_0 = t-0 = t\) and
  • \(\Delta v = v - v_0\), so that
\[ a = \frac{ v - v_0}{t} \implies v = v_0 + a t \]

This equation can be obtained by integrating the equation \(v=v_0 + a t\), where \(v=\frac{dx}{dt}\):

\[\begin{align} \frac{dx}{dt} &= v_0 + a t \\ \int_{x_0}^{x} dx &= \int_{0}^t (v_0 + a t)\ dt \\ x-x_0 &= v_0 t + \frac{1}{2} a t^2 \end{align}\]

Pay close attention to the limits of integration: the position limits go from \(x_0\) (initial position) to \(x\) (final position) as the time limits go from \(0\) (initial time) to \(t\) (final time).

This equation can be derived using the previous two equations and using the notation \(\Delta x = x - x_0\). Start by writing the second equation as:

\[\begin{align} \Delta x &= v_0 t + \frac{1}{2} a t^2 \end{align}\]

and make use of \(t = \frac{v-v_0}{a}\) from the first equation:

\[ \begin{align} \Delta x &= v_0 \left(\frac{v-v_0}{a}\right) + \frac{1}{2} a \left(\frac{v-v_0}{a}\right)^2 \end{align}\]

and a few lines of algebra to get the desired equation: \(v^2 = v_0^2 + 2 a \ \Delta x\).

Example problems using equations of motion

If you move at an average speed of \(5\ m/s\), what distance do you cover in \(10\ s\)?

Given quantities are:

\[ \begin{align} x_0 &= 0\ m, \\ \bar{v} &= 5\ m/s \\ t &= 10\ s \end{align} \]

Distance covered is \(x-x_0 = \bar{v} t = (5\ m/s)(10\ s) = 50\ m\).

A particle moves to the right with an initial velocity of \(6.0\ m/s\) and is experiencing a constant acceleration of \(-2 \ m/s^2\).

Given quantities are:

\[\begin{align} v_0 &= 6\ m/s \\ a &= -2\ m/s^2 \end{align} \]

(a) At what time does the particle stop and begin to move left?

We want to find the time \(t\) at which the particle momentarily stops \(v=0\). Therefore,

\[ \begin{align} v &= v_0 + a t \\ 0 &= (6\ m/s) + (-2\ m/s^2) t \\ \implies t &= 3\ s \end{align} \]

(b) What is the maximum displacement to the right?

We want to find \(x\) at times \(t=3\ s\) if \(x_0 = 0\).

\[ \begin{align} x &= x_0 + v_0 t + \frac{1}{2} a t^2 \\ &= 0 + (6\ m/s)(3\ s) + \frac{1}{2}(-2\ m/s^2)(3\ s)^2 \\ &= 9\ m \end{align} \]

(c) what is the displacement after \(10\ s\)?

The displacement after \(t=10\ s\) similarly is:

\[ \begin{align} x &= x_0 + v_0 t + \frac{1}{2} a t^2 \\ &= 0 + (6\ m/s)(10\ s) + \frac{1}{2}(-2\ m/s^2)(10\ s)^2 \\ &= -40\ m \end{align} \]

If you throw a ball vertically up at an initial velocity of \(10\ m/s\), how high does it go?

Given and known quantities are:

\[\begin{align} v_0 &= 10\ m/s \\ x_0 &= 0 \\ a &= -9.8\ m/s^2 \\ v &= 0 \end{align}\]

where \(v=0\) is used because we know that the ball momentarily stops at the maximum position \(x=h\). We want to find \(h\). Using one of the equations of motions:

\[ \begin{align} v^2 &= v_0^2 + 2 a (x- x_0) \\ 0 &= (10\ m/s)^2 + 2 (-9.8\ m/s^2) (h) \\ h &= \frac{100}{2 (9.8)} m = 5.1\ m \end{align} \]