In this article, we are going to discuss how to find final velocity without acceleration along with some examples and facts.

**The final velocity of the object is based on its initial velocity, the energy associated with it, position, a force acting on the object, and the time duration for the displacement.**

**Velocity**

**The velocity is defined as the ratio of displacement of the object upon the time interval given by the relation**

** Velocity=Displacement/Time**

The velocity of the object can be calculated by measuring the total displacement of the object in a specific time interval.

**Final velocity**

**Final velocity comes into the picture when the body has attained the maximum acceleration over a period of time. The acceleration is a difference between the final and the initial velocity of the object during the time.**

Based on the motion of the body, whether it is in a planar motion, uniform circular motion, or in a projectile motion, the final velocity attained by the object can be calculated.

**Final velocity of an Object in a Linear Motion**

The object moving in a plane undergoes various external forces hence the velocity of the object may not be constant every time. The final velocity of the body does depend upon the initial velocity and how much is the velocity varying with time.

**Calculate the Final velocity of an Object in a Linear Motion**

Let us see the graph of velocity v/s time of the object accelerating in a uniform linear motion with initial velocity ‘u’ and chasing the final velocity of ‘v’.

The uniformly accelerating object, the initial velocity of the object at a time t=0 is ‘u’. At time t, the velocity of the object increases to ‘v’, hence the acceleration of the object is equal to (v-u).

To measure the area of the plot in the above figure, the total area is equal to the sum of the area of the triangle (∆ABC) and the quadrilateral (□ACDO).

Since,

v=x/t

x=vt

x = Ar(∆ABC) + Ar(□ACDO)

=1/2 bh+lb

=1/2 t * (v-u)+ut

Since we are interested in finding the velocity without considering the acceleration term which is (v-u)

x=1/2 vt-1/2 ut+ut

x=1/2 vt+1/2 ut

2x=(v+u)t

2x/t=(v+u)

Therefore the final velocity of the object is

v=2x/t-u

On knowing the displacement of the object, the time taken for the displacement, and its initial velocity we can find out the final velocity picked up by the object.

Let us illustrate this with a simple example. Consider a car moving with a velocity of 20km/h starting from point A to reach point B. A car covers a distance of 60kms in 2 hours. What must be the final velocity of the car?

We know the initial velocity of the car u=20km/h,

Duration= 2hrs=120seconds

distance = 60 km

Using the formula derived above

v=2x/t-u=2*60/2-20=60-20=40km/h

Therefore the final velocity of the car will be 40kms/hr.

## Projectile motion

**An object in a projectile motion will lapse its path in a parabola. The initial and final velocity of the object will differ but the energy is conserved in a process. **Initially, when the object is on the ground, it has more potential energy which is converted into kinetic energy for its flight.

Once reaching a particular height where all of its potential energy is transformed into kinetic energy, it falls freely on the ground converting this kinetic energy into potential energy. Hence the energy is conserved in a projectile motion of the object. That is, **the sum of the kinetic and potential energy of the object before attaining maximum height is equal to the total energy after a flight.**

If ‘u’ is the initial velocity and ‘v’ is the final velocity of the object of mass ‘m’, and h_{0} is the initial height of the object from the ground and h is the highest height attained by the object in the air, then

K.E_{initial}+P.E_{initial}=K.E_{final}+P.E_{final}

1/2 mu^{2}+mgh_{0}=1/2 mv^{2}+mgh_{1}

Solving this equation further,

u^{2}+2gh0=v^{2}+2gh_{1}

v^{2}=u^{2}+2g(h_{0}-h_{1})

v^{2}=u^{2}-2g(h_{1}-h_{0})

Therefore the final velocity of the object in a projectile motion before it reaches the ground is

v=√u^{2}-2g(h_{1}-h_{0})

The change in the velocity of the object in a projectile motion is **Δ** v=v-u.

Ponder upon the helicopter dropping food parcels to people in a flood-affected area. What will be the velocity of the food parcels dropped from the helicopter flying above at the height of 600m?

Of course, the initial velocity of the parcel will be zero before dropping it from the helicopter, i.e. u=0, and the height of the helicopter from above the ground is given h=600m. Let v be the final velocity of the food parcel when it is released from the helicopter.

Substituting in the equation below

v=√u^{2}-2g(h_{1}-h_{0})

v=√0^{2}-2*10* (0-600)

v=√12000=109.54 m/s

Hence, t=600/109.54=5.47 seconds is the time required to reach the food parcel to the ground once it is dropped from the helicopter.

**Read more on Projectile Motion.**

**Velocity of the object in a circular motion**

**An object moving in a circular motion exerts a centrifugal force and centripetal force which are equal and opposite in direction and is given by the relation**

**F _{c}=mv^{2}/r**

The velocity of the object is always perpendicular to both these forces directing outward from the circular path. Due to which the velocity is the change in displacement with respect to time.

If the initial velocity of the object of mass ‘m’ accelerating in the circular track of radius ‘r’ is ‘u’, and ‘v’ is a final velocity of the object, then the net force on the object is

F=F_{2}+F_{2}

=mv^{2}/r+mu^{2}/r

=m/r ( v^{2}+u^{2})

(r/m) F=v^{2}-u^{2}

v^{2}=u^{2}+r/m F

Therefore the final velocity of the object accelerating in a circular path is

v=√ u^{2}+r/m F

**Read more on Instantaneous Velocity Vs Velocity: Comparative Analysis.**

**Frequently Asked Questions**

## Q1.**What is the final velocity of the ball accelerating downward on rising at the height of 5m from above the ground, if the mass of the ball is 500 grams? Consider the initial velocity of the ball to be 3m/s.**

Given: m=500 grams

h_{0}=5m

h_{1}=0

The initial velocity of the ball u=3m/s

Since the motion of the ball is in a projectile motion, the final velocity of the ball is

v=√ u^{2}-2g(h_{1}-h_{0})

v=√{3^{2}-2*10* (0-5)

v=√{9+100}

v=√109

v=10.44 m/s

It is evident that the speed of the ball accelerating down the ground increases due to the gravitational pull of the Earth on the objects surrounding it.

## Q2.**If an object moving with its initial velocity of 3 m/s suddenly accelerates and picks up the velocity of 10 m/s. How much distance will the object covers in 5 minutes?**

The initial velocity of the object is u=3m/s

The final velocity of the object is v=10m/s

Duration t= 5 minutes= 5* 60=300 seconds

v=2x/t-u

10=2x/300-3

13*300=2x

2x=3900

therefore x=1950 m

x=1.95 km

In a duration of 5 minutes, the object will cover a distance of 1.95 km.

## Q3.**Distance from Ratan’s house to her school is 800mtr. She starts from her house to walk to school at 7:45 with her initial velocity of 0.8m/s. She has to be present at school 5 minutes before 8 o’clock, hence she speeds up to school and reaches on time. What was her final walking speed?**

Given: d=800m,

t=10 min = 10*60 =600seconds

Initial walking speed u=0.8 m/s

Hence,

v=2x/t-u

v=2*800/600-0.8

v=8/3-0.8

v=8-2.4/3=5.6/3=1.87 m/s

Hence, the final walking speed of Ratan was 1.87 m/s.

## Q4.**What will be the velocity of the object of mass 30 kg moving with initial velocity 3m/s which accelerates at a rate of 4m/s on the application of force of 15N?**

The final velocity of the object is equal to the sum of the initial velocity and acceleration with time.

**Hence, the final velocity of the object is V _{initial}+V_{accelerating}= 3m/s+4m/s=7m/s **