Motion and Time: Meaning, Important Terms and Equations

We often use the term motion in our day-to-day life, like a car on the road is in motion. But technically, how is motion defined, and how is it related to time? What is the significance of time? How different parameters that help us to define motion are related to each other. Let us read the given article to understand Motion and Time better.

In this article, we will learn about what motion is and how do different parameters depend on the reference frame, but time is always independent. We will learn why the graphs such as acceleration time graph, velocity-time graph, or displacement time graph cannot be closed curves or will always go in the forward direction. We will also see how we can derive the equations of motion from these graphs.

We then discuss relative and projectile motion. We will also learn some basics about circular motion. Continue reading to know more.

Motion and Time

When the position of the body keeps on changing with respect to time, then the body is said to be in motion. It is important to note that, to define the position, we need to have a reference point to determine the body has change position. This reference point itself can be stationary or in motion. If the position of the body doesn’t change with respect to time, then it is said to be at rest.

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Time is an independent quantity. It cannot be stopped or reversed, and It cannot take negative values. Conventionally, we always take time to be at the x-axis; our graph cannot extend towards the left side of the y-axis as time cannot be negative. As we said that time could not be reversed or stopped; therefore, the graph will not have two values of any parameter for a body at the same instance, and thus graphs cannot be a closed curve and will always move in the forward direction.

Terms Related to Motion

Position: It is defined as the location of the particle with respect to the chosen reference point. It is denoted by \(\vec r\).

Displacement: It is defined as the change in position of the particle with respect to the reference point. We calculate it by taking the difference between the final and the initial position. It is denoted by \(\vec S\).

Distance: It is defined as the total length of the path taken by the particle.

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Velocity: The rate of change of displacement with respect to time is called velocity. It is denoted by \(\vec v\) or \(\vec u\).

Speed: The rate of change of distance with respect to time is known as speed.

Acceleration: The rate of change of velocity with respect to time is called acceleration. It is denoted by \(\vec a\).

Negative acceleration is also sometimes referred to as retardation.

Displacement, velocity, and acceleration are vector quantity while speed and distance are scalar quantities.

We also defined the average and instantaneous values of these parameters.

Instantaneous values are measure at a particular instance, while the average values are considered over a period of time.

Example: The average velocity of a body moving in a circular path in one revolution will be zero as the initial and final positions are the same. Still, the instantaneous velocity of that particle may or may not be zero.

Equations of Motion

Initial velocity is denoted by \(\vec u\).

The final velocity is denoted by \(\vec v\).

Displacement is denoted by \(\vec s\).

Acceleration is denoted by \(\vec s\).

It is important to note that acceleration is considered to be constant. If the acceleration is not constant, the equation of motions will not hold true.

The first equation of motion gives the relation between the initial velocity, final velocity, acceleration, and the time interval. It is given as,

\(\vec v = \vec u + \vec at\)

The second equation of motion gives the relation between the displacement, initial velocity, acceleration, and the time interval. It is given as,

\(\vec s = \vec ut + \frac{1}{2}\vec a{t^2}\)

The third equation of motion gives us the relation between the final and the initial velocities, acceleration, and displacement. It is given as,

\({v^2} – {u^2} = 2\overrightarrow a \cdot \overrightarrow s \)

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Position-Time Graph

At rest,

Since the particle is at rest, there will be no change in the position. Thus the graph will be a horizontal line parallel to the \(x\)-axis.

At constant velocity,

Since the velocity is constant, the slope of the graph will be constant as velocity is given by the slope of the position-time graph.

Positive constant acceleration

The graph will increase and will be parabolic. Also, the curvature gives the sign of acceleration. In the case of positive acceleration, the velocity keeps on increasing; therefore, the slope also keeps on increasing, and thus, the graph is concave upwards.

Negative constant acceleration

In the case of negative acceleration, we see that the slope of the graph keeps on decreasing, and thus the graph will be concave downwards. Also, in the given graph, the displacement is increasing, which means the initial velocity was positive.

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Velocity-Time Graph

At rest,

At rest, the velocity will be zero.

At constant velocity,

At constant velocity, the graph will be a horizontal line parallel to the \(x\)-axis.

Positive constant acceleration

The slope of the velocity-time graph gives us the acceleration; therefore, for constant positive acceleration, the graph will be linearly increasing.

Negative constant acceleration

The slope of the velocity-time graph gives us the acceleration; therefore, for constant negative acceleration, the graph will be linearly decreasing.

Acceleration-Time Graph

At rest,

The acceleration will be zero.

At constant velocity,

The acceleration will be zero.

Positive constant acceleration

For constant positive acceleration, the graph will be a horizontal line parallel to the \(x\)-axis but above the \(x\)-axis.

Negative constant acceleration

For constant negative acceleration, the graph will be a horizontal line parallel to the \(x\)-axis but above the \(x\)-axis.

Derivation of the Equations of Motion

According to the definition, acceleration is the rate of change of velocity with respect to time and in the given velocity-time graph that is equal to the slope of the graph.

\(a = \frac{{dv}}{{dt}} = \) slope of the graph

\( \Rightarrow a = \frac{{v – u}}{{t – 0}}\)

\( \Rightarrow v = u + at\)

The area under the velocity-time graph gives us the displacement of the particle.

In the given graph, the total area under the graph is given by the sum of the area of the rectangle and the area of the triangle.

\(S = ar[ACDO] + ar[ABC]\)

\( \Rightarrow s = ut + \frac{1}{2}(v – u)t\)

Now from the first equation of motion,

\((v – u) = at\)

Substituting this value, we get,

\(s = ut + \frac{1}{2}a{t^2}\)

If in the given equation we substitute the value of ‘t’ from the first equation of motion, that is,

\(t = \frac{{v – u}}{a}\)

we get,

\( \Rightarrow s = \frac{{u(v – u)}}{a} + \frac{{{{(v – u)}^2}}}{{2a}}\)

On solving the above equation, we get,

\( \Rightarrow {v^2} – {u^2} = 2as\)

Solved Problems on Motion and Time

Q.1. A particle moves in a circle with a diameter of \(14\,{\rm{m}}\) with some constant speed and completes the circular path in \(20\) seconds. Find the average velocity and average speed from \(t = 0\) to \(t = 10\,\sec \)
Ans: Given,
The diameter of the circular path is \(14\,{\rm{m}}\).
Time taken to complete one revolution is \(20\) seconds.
The average speed is given by the total distance divided by the total time taken.
\({\rm{speed}} = \frac{{{\rm{ distance }}}}{{{\rm{time}}\,{\rm{taken}}}}\)
\({\rm{speed}} = \frac{{2\pi r}}{{20 – 0}}\)
\(v = \frac{{2\pi r}}{{20 – 0}}\)
\( \Rightarrow v = \frac{{2 \times \frac{{22}}{7} \times 7}}{{20 – 0}} = 2.2\;{\rm{m}}\;{{\rm{s}}^{ – 1}}\)
Since it is given in the question that the speed is constant, therefore, in any time interval, the average speed will be equal to the average speed.
Average velocity will be given by total displacement divided by the total time taken,
In \(22\) seconds, the particle completes one revolution, so in \(11\) seconds, it will complete half the revolution; therefore, the displacement will be equal to the diameter.
Thus, the velocity will be,
\(v = \frac{d}{t}\)
\( \Rightarrow v = \frac{{14}}{{10}} = 1.4\;{\rm{m}}\;{{\rm{s}}^{ – 1}}\)

Q.2. A particle is moving with retardation of \(5\;{\rm{m}}\). The initial velocity of the particle is \(20{\rm{ m}}/{\rm{s}}\), then find the distance travelled before it comes to rest.
Ans: Given,
The initial velocity of the particle is \(20{\rm{ m}}/{\rm{s}}\).
The acceleration of the particle is \( – 5\;{\rm{m}}\;{{\rm{s}}^{ – 2}}\).
Since the particle comes to rest, the final velocity will be zero.
From the third equation of motion we have,
\({v^2} – {u^2} = 2as\)
Putting in the values, we get,
\({0^2} – {20^2} = 2( – 5)s\)
\( \Rightarrow s = 40\;{\rm{m}}\)

Summary

In this article, we discussed motion, how it is defined, and the significance of reference points. We also learned the definition of some important terms in motion. We learned about the three equations of motion. We discussed the position-time, velocity-time, and acceleration time graphs in different situations. We also derived the three equations of motion using the velocity-time graph for constant acceleration. We learned that the limitation of these equations of motion is that the acceleration should be constant.

Frequently Asked Questions

Q.1. What is a reference point?
Ans: The reference point is the point with respect to which the position of any particle is defined. Reference can be a stationary point, or it can be a non-stationary point.

Q.2. Can displacement be greater than the distance travelled?
Ans: No, displacement cannot be greater than the distance as from the definition, displacement is the distance between the final and the initial position, whereas the length of the total path travelled is the distance travelled. Example: For a particle moving in a circle, the distance travelled will keep on increasing and can be greater than the circumference, and the diameter, whereas displacement, cannot be greater than the diameter of the circle.

Q.3. Are the equations of motion always valid?
Ans: No, equations of motion are not always valid. The equations of motion hold true only when the acceleration is constant, and if the acceleration is not constant, the equations of motion will give a false result.

Q.4. Why can’t we have a position-time graph, velocity-time graph, or acceleration time graph of a particle to be a closed curve?
Ans: The position-time graph, velocity-time graph, or acceleration time graph of a particle cannot be a closed curve because if we draw a line parallel to the y-axis, then it will cut the curve at more than one point, this will signify that a particle has two values of the same parameter at the same instance which is not possible therefore the position-time graph, velocity-time graph or acceleration time graph of a particle will only go forward.

Q.5. Can velocity be greater than the speed?
Ans: No, velocity cannot be greater than the speed of the particle. It will always be equal to or less than the speed.

NCERT Solutions For Chapter: Motion And Time

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