The term velocity is used, when we want to describe in which direction an object is moving. It is defined as the distance travelled in a given direction per unit time. A moving object does not necessary maintain at a uniform velocity at all the time. It may slow down or speed up. The velocity of a body can change in two different ways, if there is a change in the speed of the body or if there is a change in the direction of its motion. When ever the velocity of an object is changing it has an acceleration. The known values of initial velocity, final velocity and time taken will help us to calculate the acceleration of any object moving in a straight line. In this present section we will learn more about final velocity.

The final velocity of the substance is the velocity of the substance that will be moving at time t. The initial velocity of a substance is the velocity it will be moving at time t = 0.
The equations of motion gives the relation between the initial velocity, final velocity, time taken , distance covered and the acceleration.
Consider an object having initial velocity '$u$' and travelling with an acceleration '$a$', after a time interval $t$ the final velocity of the object become '$v$'.
From acceleration definition, we have
$Acceleration$ = $\frac{Change\ in\ velocity}{Time\ taken}$
= $\frac{Final\ velocity – Initial\ velocity}{Time\ taken}$
$a$ = $\frac{vu}{t}$
$at = v – u$
$v$ = $u + at$
i.e., $Final\ velocity = Initial\ velocity + (Acceleration \times Time)$
If initial velocity $u$ and uniform acceleration $a$ are known.The above equation can be used to find out the final velocity $v$ acquired by a body in time $t$,
$v^{2} = u^{2} + 2as$
This equation gives the final velocity acquired by a body in covering a distance $s$.
$Average\ velocity$ = $\frac{Final\ velocity + initial\ velocity}{2}$
$Average\ velocity$ = $\frac{v + u}{2}$
The following are the final velocity problems. Solved Examples
Question 1: A static object takes 5 seconds to reach an acceleration of 10 ms^{2}. What is its final velocity?
Solution:
Solution:
Given,
Acceleration, $a$ = $10$ ms^{2}
Time taken $t$ = $5$ sec
Initial velocity $u = 0$
we have,
$Final\ velocity = Initial\ velocity + (Acceleration \times Time)$
$v$ = $u + at$
$v$ = $0 + 10 \times 5$
$Final\ velocity = 50$ms^{1}
Question 2: If a car can accelerate uniformly at 2.5 ms^{2 }and start from a velocity of 36 km/h, find the velocity after 8 sec?
Solution:
Solution:
Given,
Initial velocity $u$ = $36$ km/h = $\frac{36 \times 1000m}{60 \times 60s}$ = $10 m/s$
Initial velocity $u$ = $36$ km/h = $\frac{36 \times 1000m}{60 \times 60s}$ = $10 m/s$
Acceleration $a$ = 2.5 ms^{2}
Time taken $t$ = $8$ sec
we have,
$Final\ velocity = Initial\ velocity + (Acceleration \times Time)$
$v$ = $u + at$
$v$ = $10 + 2.5 \times 8$
$Final\ velocity = 30$ms^{1}
The final velocity of the car is 30 ms^{1}
Question 3: A Stone is dropped freely from a height of 100 m. With what velocity will it hit the ground? Neglect the air resistance while calculating.
Solution:
Solution:
Given,
$s = 100$m
$u = 0$
$a = g = 9.8$m/s
we have
$v^{2} – u^{2} = 2as$,
Hence, $v^{2} = u^{2} + 2as$
$v^{2} = 0 + 2 \times 9.8 \times 100$
$v^{2} = 1960$ ms^{2}
or
$v = 44.28$ m/s.
The stone will hit the ground with a velocity of $44.28$ m/s.