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# Kinematic Equations Made Easy

I don’t feel that I should explain the importance of kinematics in Physics. **Kinematic equations** form the very foundation of any question you intend to solve in physics. Be it uniform rectilinear motion or rotational motion, these equations will always help you to find the correct answer.

Whether you are preparing for NEET or JEE you must have a strong command in kinematics if you want to crack these types of competitive exams. Here in this article, I have tried to strengthen your concepts by explaining all types of kinematic equations.

## What are Kinematic Equations?

Kinematic equations are equations showing the dependence of the main kinematic characteristics (radius vector, coordinates, velocity, acceleration) on time.

## Basics of Kinematics

In mechanics, we will use five basic SI units:

Measurement | Unit | Symbol |
---|---|---|

Mass | Kilogram | kg |

Length | Meter | m |

Time | Second | s |

Angle | Radian | rad |

Solid Angle | Steradian | cf |

The quantities used in physics are of two types:

**Scalar Quantity**— A scalar is a value characterized by a numerical value (it can be positive or negative).**Example:**Speed**Vector Quantity**— A vector is a quantity characterized by both a numerical value (modulus of a vector, a positive number) and a direction.**Example:**Velocity

There are five **Kinematic Variables** that link any type of kinematic equation. They are:

Displacement | Δx |

Initial velocity | v or _{0}u |

Final velocity | v or v_{f} |

Time interval | t |

Constant acceleration | a |

## Kinematic Equations

These can be grouped into Rectilinear Kinematics equations for linear motion and Rotational Kinematics Equations for angular motion. Let’s have a look:

### Basic Kinematic Equations for Linear Motion

The translational or linear motion of the body is the one in which all its points move along the same trajectories and at any given moment they have equal speeds and equal accelerations. There are four basic equations of kinematics for linear or translational motion. These are:

$$v=v_0+at$$

$$Δx=t(v+v_0)/2$$

$$Δx=v_0t+\dfrac{1}{2}at^2$$

$$v^2=v_0^2+2 a Δx$$

Questions are frequently asked based on this formula. Do use it and thank me later!

**Remember: These equations are only applicable when there is uniform motion, and acceleration and velocity, both are constant.**

Let me give you one example:

Suppose a car is moving with an initial velocity of 20m/s and comes to rest in 5 seconds. You are asked to find acceleration and displacement covered.

**Solution:** According to the question the initial velocity , u =20m/s

Since the body is coming to rest that means the final velocity, v= 0 m/s

Time taken by the body to come to rest, t=5 sec

So, applying the first equation of motion we get, v= u +at, i.e 0=20 + 5a

thus, acceleration, **a = $-4m/s^2$**

Now, since the body is coming to rest after time, t that means there is some retarding force that is being applies to the body to to which is comes to rest , since the acceleration is negative that means the retardation is occuring against the direction of motion.

Now, to find the **displacement** we can use second equation , so

$Δx=t(v+v_0)/2$

thus, Δx= 5(20+0)/2 = 50m.

**Also read:** Tunnel Through the Earth

### Rotational Kinematics Equation for Angular Motions

Rotational motion is a movement in which all points of the body move in circles, and the centers of all the points lie on one straight line – the axis of rotation. There are few changes when compared to linear equations of motion.

- Displacement is replaced by a change in angle and it is denoted by theta (Θ)
- Velocities are replaced with angular velocities
- Acceleration is replaced by angular acceleration
- Time remains constant

$$ω=ω_0+αt$$

$$Θ=1/2(ω+ω_0)t$$

$$Θ=ω_0t+\dfrac{1}{2}αt^2$$

$$ω^2=ω^2_0+2αΘ$$

where ω is the final angular velocity, $ω_0$ is the initial angular velocity, t is time and Θ is displacement and α is angular acceleration.

## Problem-solving strategy for Kinematics

Let us now understand how to use the above equations to obtain more details about the motion of the object in question. For this problem-solving strategy, we’ll have to follow the steps given below:

- Put together a detailed diagram of the physical situation.
- Identify the given information and proceed to list it in variable form.
- Identify the unknown information and proceed to list it in variable form.
- Identify and list the equation that you need to use to obtain unknown information from known information.
- Substitute known values into the above equation and solve for the unknown information using the required algebraic steps.
- Verify your answer and make sure that it is mathematically correct and reasonable.

We will now understand how to use this strategy by solving two different example problems down below.

### Example 1

**Ross is ****driving ****with a velocity of +30 m/s**** and coming closer to a stoplight****. ****The light turns yellow, and Ross skids to a stop after applying the brakes. ****If his acceleration during this process is -8.00 m/s**^{2}**, then calculate the displacement of his car during the**** skidding process.**

First of all, notice that the direction of the velocity and the acceleration vectors have been denoted by a + and a – sign respectively. In order to solve this problem, you must draw a schematic diagram of the situation described, as shown below.

After that, you need to identify and list the known information in variable form.

In this case, we can infer the value of v to be 0 m/s because Ross’ car skids to a stop. On the other hand, the initial velocity of the car, u, is 30 m/s and its acceleration is given as -8.00 m/s^{2}. Do not forget to give due attention to the + and – signs for the concerned quantities; failing to do so can lead to erroneous calculations.

After this, you need to list the unknown or desired information in variable form. In the given problem, you require information about the displacement of Ross’ car. Thus, S is the unknown quantity we’re after. You now need to identify a kinematic equation that will help you determine this quantity.

We’ve already studied the four kinematic equations above. Generally, you always have to select the equation that contains the one unknown and three known variables. In this example, the unknown variable is S and the three known variables are v, u, and a. Thus, you must find an equation that has these four variables listed in it.

After looking closely at each of the four kinematic equations above, you will notice that the second equation contains all the four variables:

v^{2} = u^{2} + 2aS

After identifying the equation and writing it down, you need to substitute the known values into the same and solve for the unknown information using proper algebraic steps. I’ve described this step down below:

(0 m/s)^{2} = (30.0 m/s)^{2} + 2 × (-8.00 m/s^{2}) × S

0 m^{2}/s^{2} = 900 m^{2}/s^{2} + (-16.0 m/s^{2}) × S

(16.0 m/s^{2}) × S = 900 m^{2}/s^{2} – 0 m^{2}/s^{2}

(16.0 m/s^{2}) × S = 900 m^{2}/s^{2}

S = (900 m^{2}/s^{2})/(16.0 m/s^{2})

S = (900 m^{2}/s^{2})/(16.0 m/s^{2})

**S = 56.3 m**

Thus, upon solving for S and rounding the answer to the third digit, we can see that Ross’ car will skid a distance of 56.3 meters. Finally, you need to check the answer to make sure that it is both accurate and reasonable.

The value of the displacement does sound reasonable enough, because it will obviously take a car quite a bit of a distance to skid from 30.0 m/s to a halt. After all, 56.3 m is roughly half the length of a football field.

To check for accuracy, we need to substitute the calculated value back into the equation for displacement and check if the left and right sides of the equation are equal to each other. That is indeed the case here, so you can be assured that the answer is correct.

### Example 2

**Stan is waiting at a stoplight that soon turns green.**** He now proceeds to accelerate from rest at a rate of 6.00 m/s**^{2}** for a time of 4.10 seconds.**** Calculate the displacement of Stan’s car during this time period.**

Like the previous example, the solution to this problem starts with us putting together a detailed diagram of the situation described, as shown below.

After that, we must identify and list the known information in variable form. In this case, we can infer the value of v to be 0 m/s because the car is initially at rest. Similarly, the acceleration a is given as 6.00 m/s^{2} and the time t is 4.10 seconds.

We must now list the unknown or desired information in variable form. Here, the problem requires you to calculate the displacement of the car; therefore, S is the unknown information. The next step of our strategy is to identify a suitable kinematic equation to solve for the value of S.

Inspecting the four kinematic equations given above, you can see that the first equation contains all the four variables we’re dealing with:

S = ut + ½ at^{2}

Now that we’ve identified the equation, we can substitute the known values into the same and use proper algebraic steps to solve for S, as shown below:

S = (0 m/s) × (4.1 s) + ½ × (6.00 m/s^{2}) × (4.10 s)^{2}

S = (0 m) + ½ × (6.00 m/s^{2}) × (16.81 s^{2})

S = 0 m + 50.43 m

**S = 50.4 m**

Rounding the calculated value of S to the third digit, we ascertain that Stan’s car will travel a distance of 50.4 meters.

Finally, we need to check the answer to make sure that it is both reasonable and accurate. It seems reasonable enough that a car accelerating at a speed of 6.00 m/s^{2}^{ }will attain a speed of about 24 m/s in 4.10 seconds. The distance over which the car would be displaced during this time period should be about half the length of a football field. Thus, the value of S we’ve calculated is quite reasonable.

To check for accuracy, you need to substitute the calculated value of S back into the equation for displacement, and see whether the left side of the equation is equal to the right side or not. Doing so, you will find out that it holds true in this case. Thus, our calculation is accurate.

With the help of the two example problems above, you can learn how to combine kinematic equations with a handy problem-solving strategy to determine the unknown parameters of motion for a moving entity. If you know any three of the parameters, you can find out the fourth one.

We’ll now understand how to apply this strategy to deal with free fall situations.

## Kinematic equations and free fall

A free-falling object is an object that is falling solely under the influence of gravity. In other words, any body that is moving and being subjected only to the force of gravity is said to be in a state of **free fall**. A body falling in this manner will experience a downward acceleration of 9.8 m/s^{2}. This holds true whether the object in question is falling downwards or rising upward towards its peak.

Like any other moving object, we can describe the motion of a body in free fall using the four kinematic equations we’ve already studied.

### Applying the concept of free fall to solving problems

When we’re using the kinematic equations to analyze the free fall of a body, we need to consider certain conceptual characteristics of free fall:

A body in free fall experiences an acceleration of -9.8 m/s^{2}, where the – sign indicates a downward acceleration. Thus, we should take the value of a as -9.8 m/s^{2}^{ }for any freely falling object whether it’s stated explicitly in the problem or not.

If an object is simply dropped (not thrown) from a height, then its initial velocity u will be 0 m/s.

If an object is projected upwards vertically, then it will progressively slow down as it rises upwards. At the instant when it reaches the peak of its trajectory, its velocity is 0 m/s. We can use this value as one of the parameters of motion in the kinematic equations. For example, the final velocity v after travelling to the peak will have a value of 0 m/s.

In the above situation, the velocity at which the body is project is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. For example, an object thrown vertically upwards with a velocity of +20 m/s will have a downward velocity of -20 m/s after returning to the same height.

We can combine these concepts and the four kinematic equations to efficiently solve problems that deal with the motion of freely falling objects. Given below are two problems to illustrate the application of the concepts principles of free fall to solve kinematic problems. In both of them, we’ll use the problem-solving strategy described earlier.

### Example 3

**A boy drops a ball from the top of a roof located 8.52 ****meter****s above the ground. Calculate the time required for the ball to reach the ground.**

To solve the problem, we must begin by drawing a schematic diagram of the given situation, as shown below.

After that, we need to identify and list the known information in variable form. In this case, the problem gives us only one piece of numerical information – the height of the roof above the ground (8.52 meters). This indicates that the displacement S of the ball is -8.52 m, the – sign indicating that the displacement is downward.

Based upon our understanding of the concepts of free fall, we must obtain the remaining information from the problem statement. For example, we can infer the initial velocity u to be 0 m/s since the boy drops the ball from rest. Since the ball is falling freely, the acceleration a is inferred to be -9.8 m/s^{2}. After that, we must list the unknown or desired information in variable form. In this case, it is the time of fall t.

Now, we need to identify a suitable kinematic equation to calculate the unknown quantity. In this problem, the first equation contains all the four variables we’re dealing with:

S = ut + ½ at^{2}

We can now substitute known values into this equation and solve for t using proper algebraic steps, as shown below:

-8.52 m = 90 m/s) ×(t) + ½ × (-9.8 m/s^{2}) × (t)^{2}

-8.52 m = (0 m) × (t) + (-4.9 m/s^{2}) × (t)^{2}

-8.52 m = (-4.9 m/s^{2}) × (t)^{2}

(-8.52 m)/(-4.9 m/s^{2}) = (t)^{2}

1.739s^{2} = t^{2}

**t = 1.32 s**

After rounding the value of t to the third digit, we can see that the ball will fall for a time of 1.32 seconds before it lands on the ground. Finally, we must make sure that our calculated value of t is both reasonable and accurate. Since the ball is falling down a distance of about 10 yards, it seems reasonable enough that the time taken is between 1 and 2 seconds. Like before, you can substitute the calculated value of t back into the equation and see that both sides of the equation are identical.

### Example 4

**Tom throws his toy airplane vertically upwards with an initial velocity of 26.2 m/s. ****Calculate the height to which the toy will rise above its initial height.**

As always, we must begin with a schematic diagram of the given situation as shown below.

Now, let’s identify and list the known information in variable form. This problem explicitly states only one piece of numerical information – the initial velocity u of the vase is +26.2 m/s. The + sign, of course, indicates the upward direction of u.

We must obtain the remaining information from the problem based upon our understanding of the concepts of free fall. We can infer the value of the final velocity v to be 0 m/s because the final state of the toy is the peak of its trajectory. Similarly, its acceleration a is -9.8 m/s^{2}.

After this, we need to list the unknown or required information in variable form. In this case, the displacement S of the toy is the required information. Going through the four kinematic equations, you can see that the second equation contains all the four variables we’re dealing with:

v^{2} = u^{2} + 2aS

We can now proceed to substitute the known values into the equation and solve for S using appropriate algebraic steps, as shown below:

(0 m/s)^{2} = (26.2 m/s)^{2} + 2 × (-9.8 m/s^{2}) × S

0 m^{2}/s^{2} = 686.44 m^{2}/s^{2} + (-19.6 m/s^{2}) × S

(-19.6 m/s^{2}) × S = 0 m^{2}/s^{2} – 686.44 m^{2}/s^{2}

(-19.6 m/s^{2}) × S = -686.44 m^{2}/s^{2}

S = (-686.44 m^{2}/s^{2})/(-19.6 m/s^{2})

**S ****= 35****.****0**** m**

Thus, we can see that the toy will travel upwards for a displacement of 35.0 meters before reaching its peak. Finally, we need to make sure that our answer is both reasonable and accurate. For the latter, you can substitute the calculated value of S back into the equation and see that both sides of the equation are identical.

The problem states that Tom throws the toy with a speed of 26.2 m/s, which is unlikely to make it further than around 100 meters in height. However, it will certainly make it past a minimum height of 10 meters. Thus, our answer falls within this range of reasonability.

If you know at least three parameters of motion, you can use the kinematic equations to calculate the value of an unknown motion parameter. When we are dealing with a body in free fall, we usually know the value of acceleration. In a lot of cases, we can also infer another motion parameter with a sound knowledge of certain fundamental principles of kinematics.

## Conclusion

This was all about **kinematic equations for both linear and angular motions.** I hope I was able to make your concepts a bit clearer so that you don’t face any problems while solving questions on mechanics. Do utilize the concepts and prepare well. *Happy learning!*

**References**

- https://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations-and-Problem-Solving (For example 1 & 2)
- https://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations-and-Free-Fall (For example 3 & 4)
- https://en.wikipedia.org/wiki/Kinematics
- Fundamentals of Kinematics and Dynamics of Machines and Mechanisms