What Are the Four Kinematic Equations? 

Candidates studying science subjects must have witnessed the topic of kinematic equations in Physics. While their importance is immense for setting up the base of Physics, there are many students who remain unfamiliar with the concept. But now, worry not. Here are the insights to help you solve the related problems. 

What is Kinematics? 

Kinematics is the branch of Physics that considers motion with respect to time. Kinematic equations are a group of equations comprising terms like displacement, velocity, and acceleration. These equations assist in understanding and calculating an object’s motion and position concerning time. 

Important Variables in Kinematics Equations 

The kinematics equations have the following variables in their equation: 

Terms	Symbols
Displacement	x
Initial velocity	v0 or vi
Final velocity	vf
Acceleration	a
Time	t

Kinematic Equations 

There is a set of four kinematic equations that allow the calculation of displacement and velocity. However, these can be molded to calculate the required values as per the given data. It must be noted that the value of at least three variables must be known for the calculation of the fourth variable. The linear kinematic equations are as follows: 

d=vit+1/2at2

d=vi+vi/2*t

vf2=vi2+2ad

vf=vi+at

Besides linear, there are rotational kinematic equations as well. These deal with the rotational motion of the body. In this set of equations, only certain variables are replaced. The rotational kinematics equations are as follows: 

=it+1/2 *at2

=i+i/2 *t

f2=i2+2a

f=i+at

Here, it refers to angular velocity and refers to angular position. 

Method to Solve Kinematic Equation 

Once you know all the equations and you have the question, you will go forward to solve the kinematic equation. Hence, here is a stepwise approach to solving the kinematic equation: 

Step 1: Know the given information 

With the question in front of you, you must identify the known variables accurately. While it is mostly mentioned in numerical values, it is also sometimes stated theoretically. For instance, you might encounter a statement saying, ‘The body started from rest to reach the velocity of 10 meters/second.’ The statement indicates the initial velocity will be 0 m/s.

Step 2: Recognize the missing variable 

All the kinematic equations have almost the same variables, just arranged differently. Hence, you will need to figure out the right kinematic equation to use and the missing variable. 

Step 3: Fill in the values and calculate

Since you have chosen your equation and clearly know the variables, you are now good to go. Substitute the values, and you will get the answer. You might be required to use the calculator for complex calculations. 

Examples of Kinematic Equations 

Let’s now understand kinematic equations through practical application examples. 

Example 1: Find out the final velocity of the car that is moving with the acceleration of 3.5 m/s2 for 5 seconds after being parked for 10 minutes. 

Solution: Firstly, the variables given in the question are initial velocity, final velocity and time. Now, since the car was parked, it was at zero initial velocity. As we have acceleration and time, we use the kinematics equation vf=vi+at. 

Substituting the values in the equation: 

vf=0+3.55

vf=17.5 m/s

Hence, the final velocity of the car is 17.5 m/s. 

Example 2: Find the distance travelled by a car if it moves from rest with acceleration 6.5 m/s2 for the duration of 15 seconds. 

Solution: The acceleration, initial velocity and time are given where we have to calculate the distance. Hence, the kinematics equation to be used here is: 

d=vit+1/2 * at2

Substituting the values, we get: 

d=0 * 15+1/2 * 6.5 *15(2)

d=0+1/2 * 6.5 *15(2)

d=731.25 metres

Hence, the distance travelled by the car is 731.25 meters in 15 seconds with a given acceleration. 

Example 3: A truck starts with an initial speed of 4 m/s and halts at the final speed of 16 m/s. It undergoes uniform acceleration for 10 seconds. Find the distance covered by the truck in the mentioned duration. 

Solution: We have the initial and final velocity and the magnitude of time. Thus, the kinematics equation to be used is: d=vi+vi/2 * t. 

Substituting the values, we get: 

 d=4+16/2 * 10

d=20/2 * 10

d=100 meters

Hence, the truck covered 100 meters of distance with the acceleration for 10 seconds. 

Example 4: A driver is travelling with a car with an initial speed of 14 m/s. He suddenly applies the brakes, leading the car to stop after covering 45 meters. Find the deceleration of the car. 

Solution: The initial and final velocity and distance are given. Since the car will be stopping, the final velocity will be zero. So, we will be calculating the deceleration using the kinematics equation:

vf2=vi2+2ad

Substituting the values, we get: 

02=14(2)+2 * a *45

0=196+90a

90a=-196

a=-196/90

a=-2.18 m/s2 

Since we have a negative sign associated with acceleration, it indicates deceleration, i.e. slowing down. The deceleration of the car -2.18 m/s2.

Ace Kinematic Equations With Turito 

Weren’t these kinematic equations easier to solve? Just put in the values and solve, and you get the answer. However, the level goes higher as you proceed to higher classes, boards, and competitive exams. It requires an in-depth understanding and familiarity with the derivation of the kinematics equations to solve more complex problems. 

Turito helps the students with conceptual clarity. Having assisted numerous students in acing their careers with the right guidance and mentoring, Turito ensures one-to-one attention to the students. Further, the doubt-solving sessions are provided so that all the questions that arise in the candidate’s mind are solved. 

Frequently Asked Questions on Turito