In class eleventh we study equation of motion that govern every motion in classical world. In this article, we’ll examine that how to find final velocity with acceleration and distance using kinematic equations in this article.

**We have three equations of motion having distance, initial and final velocities, time and acceleration. We can easily calculate final velocity with acceleration and distance using third equation of motion i.e. v ^{2} = u^{2} + 2as**

Where vis final velocity of object,

u is u initial velocity of object,

a is acceleration of object

And s is distance travelled by object.

**The word “Kine” mean motion and the word “Matics” mean math. So, the word kinematics mean mathematics of motion.** Kinematic equations explain very basic principle of velocity, acceleration and position of some system or body. These three equations of movement regulate the motion of a body in one – dimension, two dimensions, and three dimensions.

One of the main essential concepts in Physics is the formulation of the equations of motion. Kinematic equations are mathematical equations which describe how a material body behaves when it moves with respect to time.

**How to find final velocity with initial velocity acceleration and distance?**

Three equations of motion can be used to compute parameters like displacement(s), velocity (initial and final), time(t), and acceleration (a).

**The three equations of motion are as follows: **

**First equation of motion is v = u + at**

**Second equation of motion is** s = ut+ 1/2at^{2}

**Third equation of motion is v ^{2} = u^{2} + 2as**

We can derive these equations by using various methods like algebraic method of derivation for equation of motion, graphical method of derivation for equation of motion and calculus method for derivation of equation of motion.

**Derivation of first equation of motion by algebraic method **

By definition of acceleration, we understand it as; ** **

“Rate of change of velocity of some body or object is acceleration.”

So, mathematically we can express it as a =

a = v-u/t

Rearranging the above equation, we get

v = u+at

This is first equation of motion.

Where v is final velocity of object, u is initial velocity of object and t is time taken by object.

**Derivation of second equation of motion by algebraic method **

By definition of velocity, we understand it as;

“Rate of change of displacement of some body or object is velocity.”Velocity = Displacement/Time …………………..(3)

Reorganising this equation we get

Displacement=velocity\time

In equation number three, we can substitute velocity with average velocity if velocity is changing with time and again write the third equation as;

Displacement=(initial velocity + final velocity)/t * time

We get the following equations when we replace the representations given in the formulation of the first equation of motion with the abbreviations involved in the formulation of the second equation of motion;

s = (u+v)/2 * t …………………(6)

We can plug the value of v i.e. v=u+at from first equation of motion into equation (6), we get

s=u+(u+at)/2* t

s=2u+at/2 * t

s= (2u/2 + at/2) * t

s = (u+1/2at) * t

After simplifying more this equation become;

s = ut+1/2 at^{2}

This is second equation of motion.

**Derivation of third equation of motion **

By definition of displacement, we understand it as;

“Rate of change of position of some body or object is displacement.”

Displacement= (initial velocity + final velocity)/2* time

Replacing the standard abbreviations into above equation, we get following equation;

s = (u+v)/2 * t

From first equation of motion, we know that

v=u+at, reorganising this equation we can write it as;

t = v-u/a

Replacing the value of t into displacement formula, we get;

s = v+u/2, v-u/a

s = v^{2} – u^{2}/2a

2as = v^{2} – u^{2}

Reorganising equation (9), we get

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

This is the third equation of motion.

**Examples **

**Example 1 **

**A car achieves an average acceleration 25m/s**^{2}. Assume the car accelerates at this pace for 6 seconds from a standstill. How much displacement car will cover in this instant of time?

^{2}. Assume the car accelerates at this pace for 6 seconds from a standstill. How much displacement car will cover in this instant of time?

**Solution**

**As we can see here initial velocity of car, acceleration of car and time taken by car is given in this situation, so we can use third equation of motion to solve this example. **

**Since car is at rest in beginning, so initial velocity is zero i.e. u = 0 **

**Time taken, **t = 6 seconds

**And acceleration,** a = 25m/s^{2}** **

**To find out that how much distance the car covered, we will use second equation of motion **

s = ut+ 1/2 at^{2}

s = 0(6) + 25 m/s^{2} (6s)^{2}

s= 25 m/s^{2} 36s^{2}

s = 900m

**So, the distance covered by car is 900 meters. **

**Example 2 **

## In vertically upward direction a piece of wood is thrown and it has a velocity of 6m/s and during its motion it achieves acceleration of 12m/s^{2} in downward direction. How much height that piece can attain and how much time the piece will consume to reach that height?

**Solution**

**As we can** **see that velocity given to the piece of wood in starting is 6m/s and because at the end piece of wood will hit the ground and will be at rest so the final velocity of wooden piece will be zero. Since acceleration, initial velocity and final velocities are known, we will use third equation of motion to calculate the height achieved by piece of wood. **

**Given; **

**Initial velocity, u = 6m **

**Final velocity, u=0**

**Since a is in downward direction, it will be retarded motion and we will use negative value of acceleration. **

** Acceleration, a = -12 m/s ^{2} **

**Third equation of motion is **

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

** Putting** **given values in above equation, we get **

0^{2} = 6m/s^{2} + 2(-12m/s^{2})s

0 = 36m^{2}/s^{2} – 24m/s2 s

36m^{2}/s^{2} = (24m/s^{2})s

s = 36m^{2}s^{2}/ 24m/s^{2}

s = 1.5m

**So, the height attain by wooden piece is 1.5 meter. **

**We’ll now compute how long it took the wooden pieces to reach that altitude. **

**As we have acceleration, initial velocity, final velocities and the height attain by wooden piece, we will use first equation of motion to calculate the time taken to achieve that height. **

**Given; **

**Initial velocity, u = 6m/s**

**Final velocity, u=0**

**Acceleration, a=-12m/s ^{2}**

s = 1.5m

** First equation of motion is **

v = u+at

** Putting the above data available into this equation, we get **

0 = 6 m/s + (-12m / s^{2}) t

6m/s = (12m/s^{2} )t

**Reorganising above equation **

t = **6m/s** / 12 m/s^{2}

t = 0.5s

**So, time taken by wooden piece to attain maximum height is 0.5 seconds. **

**Problems **

**Problem 1 **

**A car starts moving with initial velocity 50m/s and during its motion it archives **10m/s^{2}** acceleration. How much will be the velocity after 10 seconds of its motion and after that calculate the distance covered by moving car after 10 seconds**.** **

**Solution**

**Here in this problem, we know initial velocity of car, acceleration of car and time taken by car. So, to find out the final velocity of car, we will use first equation of motion; **

**That is v = u+at **

**Given values are **

u = 50 m/s

a = 10 m/s^{2}

t=10 seconds

**Putting the above data available into first equation of motion, we get **

**v=50m/s+(10m/s ^{2}(10seconds)**

**v=50m**/s + 100m/s

v = 150 m/s

**SO, the final velocity of car is 150m/s.**

**Now for calculation of distance covered by car we will use third equation of motion. **

**Third equation of motion is **

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

**Putting respective values in third equation of motion, we get**

150^{2} = 50^{2} + 2(10m/s^{2}) s

22500m^{2}/s^{2} = (2500 m^{2}/s^{2} + 20m/s^{2})s

22500m^{2}/s^{2} – 2500m^{2}/s^{2} = (20 m/s^{2} )s

20000m^{2}/s^{2} = (20 m/s^{2} )s

s = 20000m^{2}/s^{2}/ 20m/s^{2}

s = 1000m

**So, the distance covered by car in 10 seconds is one thousand meters. **

**Frequently asked questions |FAQs **

**Q.** **What is kinematics? **

**Ans.** **Kinematics is a branch of classical mechanics that handles the motion of objects, things, and systems without considering the causes of movement. (i.e., forces). The “mathematics of motion” is a term used to describe the study of kinematics. **

**Q. How you will describe motion?**

**Ans.** Here are bodies in motion wherever we look. Anything from a tennis game to a space probe flyby of the planetary Neptune has movement. Your heart pumps blood via your veins as you sleep. Even in inanimate things, the oscillations of atoms and molecules keep moving.

**Fascinating movement issues might emerge, such as how long would it take a space mission to reach Mars? When a football is thrown at a specific angle, where will it land? Knowing motion, on the other hand, is essential for grasping other physics concepts. The analysis of force, for instance, necessitates comprehension of acceleration. **

Kinematics investigates the paths of dots, lines, and various geometric entities, and also its derivative qualities, to explain movement (such as velocity and acceleration). In astronomy, kinematics is used to explain the movement of astronomical planets and systems; in mechatronics, automation, and ergonomics, it is utilized to explain the movement of processes made up of connected elements.

**Q. What is the velocity of an entity?**

We can define velocity in terms of distance and time.

“Rate of change of displacement with respect to time is called velocity.”

The rate of flow of an item, as well as its heading, is described as velocity. The velocity of an object informs you how rapidly it is going and in which way it is traveling. It’s identical to speed, but with the addition of direction. Although speed is a directionless quantity, velocity is a directional quantity.