Kinematics | Definition, Graphs & Theory

• Instantaneous acceleration refers to the acceleration at a specific instant. It is a very small change in velocity at an infinitesimally small time interval.
• Average acceleration is the rate in change of velocity. It is the acceleration described and mathematically expressed above.
• Deceleration is characterized by acceleration which has a direction opposite to that of velocity. A car slowing down as it moves forward undergoes deceleration since the acceleration's direction is opposite (backward) to that of velocity (directed forward).

Fig 2: A car is decelerating when its acceleration is opposite the direction of its velocity.

Equations of Motion

There are equations that can be used when dealing with word problems involving constant acceleration. It encompasses the relation between velocity, position, and acceleration. The four equations, also known as the UAM (uniformly accelerated motion) equations are as follows:

• Equation 1: {eq}v=v_0+at {/eq}

• Equation 2: {eq}\Delta x=v_0 t +\frac{1}{2}at^2 {/eq}

• Equation 3: {eq}v^2=v_{0}^2+2a\Delta x {/eq}

• Equation 4: {eq}\Delta x =\frac{1}{2}(v+v_{0})t {/eq}

where v is the final velocity, {eq}v_{0} {/eq} is the initial velocity, t is the time, {eq}\Delta x {/eq} is the displacement, and a is the acceleration. Each equation is used based on the given and unknown quantities in a problem.

Consider the examples below that show how the equations of motions are used in problem-solving.

Example 1

A car starts at rest and has an acceleration of 5 m/s{eq}^2 {/eq}. What is its velocity after it covers a distance of 50 m?

Step 1

Identify the given quantities.

The given quantities are initial velocity, acceleration, and displacement.

{eq}v_{0}=0 \text{ m/s} {/eq}

{eq}a=5 \text{ } \frac{m}{s^2} {/eq}

{eq}\Delta x = 50 \text{ m} {/eq}

Step 2

Identify the unknown quantity.

Find the final velocity of the car.

Step 3

Choose the most appropriate equation.

The best equation to use based on the given and unknown quantities is Equation 3: {eq}v^2=v_{0}^2+2a\Delta x {/eq}

Step 4

Substitute the given values to the equation.

{eq}v^2=(0\text{ m/s})^2+2(5 \text{ } \frac{m}{s^2})(50 \text{ m}) {/eq}

Step 5

Find the answer.

{eq}v=22.4 \text { m/s} {/eq}

Thus, the final velocity of the car is 22.4 m/s.

Example 2

A car moves at a constant velocity and covers 80 m for 6.0 s. If the driver stepped on the brakes and came to a stop after 5.0 s, what is the magnitude of its acceleration?

Step 1

Identify the given quantities.

Initially, the car is moving at a constant velocity. Calculate the initial velocity using:

{eq}v_{0}=\frac{d}{t}=\frac{80 \text{ m}}{6.0 \text{ s}}=13.3 \text{ m/s} {/eq}

Both the final velocity and time are given.

{eq}v=0 \text{ m/s} {/eq}

{eq}t=5.0 \text{ s} {/eq}

Step 2

Identify the unknown quantity.

Find the magnitude of the car's acceleration.

Step 3

Choose the most appropriate equation.

The best equation to use based on the given and unknown quantities is Equation 1: {eq}v=v_0+at \rightarrow a=\frac{v-v_{0}}{t} {/eq}

Step 4

Substitute the given values to the equation.

{eq}a=\frac{0 \text{ m/s}-13.3 \text { m/s}}{5.0 \text { s}} {/eq}

Step 5

Find the answer.

{eq}a=-2.66\text { } \frac{m}{s^2} {/eq}

Thus, the magnitude of the car's acceleration is 2.66 m/s{eq}^2 {/eq}. The negative sign only indicates that the car is decelerating.

Graphical Representation

In kinematics, graphs, specifically motion graphs, are used to easily visualize an object's motion. The most common kinematic motion graphs are displacement-time and velocity-time graphs.

In a displacement vs time graph, the slope of the graph is equivalent to the velocity of the object. In Fig. 3 a few examples of displacement-time graphs and their corresponding motion descriptions are shown. Graph A illustrates a stationary object at a certain distance from a reference point. It shows a straight horizontal line. Graph B shows an object moving at a constant positive velocity, as illustrated by the straight diagonal line with a positive slope. Graph C represents an object having an increasing velocity in the positive direction, while graph D suggests that the object is moving with a positive decreasing velocity. Graphs C and D have changing gradients, as shown by the curved lines.

Fig. 3: Displacement-time motion graphs and their corresponding interpretations

The displacement-time graphs in Fig. 3 have corresponding velocity-time graphs. The area of a velocity-time graph represents the displacement of the object, while its slope describes the object's acceleration.

Fig. 4: Velocity-time motion graphs and their corresponding interpretations

In the velocity-time graphs shown, graph A illustrates an object at rest, while graph B shows an object moving at constant velocity. For both graphs the line is horizontal, indicating that the object has zero acceleration. Graph C suggests that the velocity of the object is positive and increasing at a constant rate. It forms a straight diagonal line with a positive slope, indicating that the acceleration is positive. Graph D represents an object slowing down while moving in a positive direction. It shows a straight diagonal line with a negative slope, which means that the object is decelerating.