Velocity, acceleration, and time are the fundamental quantities to derive the equation of motion. In general, the time derivative of the velocity gives the acceleration.

**In kinematics, the velocity can be found by using acceleration and time. Since velocity and acceleration are associated with magnitude and direction, in order to find out velocity, we use both the algebraic method and integral calculus. Using both methods, how to find velocity with acceleration and time is discussed in this post.**

Let us consider the body is moving with acceleration ‘a’ covering certain distance at the time ‘t.’

**By Algebraic method:**

From the kinematic definition, **acceleration is the rate of change of velocity of the traveling body.**

a=v/t

Here we consider; initially, the body possesses minimal velocity; hence the initial velocity can be considered approximately zero.

On rearranging the terms, we get the velocity of the body as;

v = a*t

**By integral calculus method:**

**The time derivative of the velocity gives the acceleration of the body**. It is given by the following equation.

d/dt[v(t)]= a(t)

Rearranging the above equation

dv(t) = a(t) dt

Integrating the above equation with respect to time t

∫d/dt[v(t)]=∫a(t) dt+C

Where; C is the integral constant.

Therefore; v = at + C

The above equation gives the velocity; thus, acceleration times the time gives the velocity.

**How to find velocity with acceleration and time graph**?

**The graph of acceleration vs. time is plotted, leading to finding out the various physical quantities like jerk and velocity. The area covered under the acceleration–time graph gives the velocity.**

For example, a car is moving with an initial velocity of 16 m/s. As with the time, the car begins to accelerate. The acceleration of the car is constant with time. After some time, the car suddenly stops, which is represented in the graph given below.

The dotted line is used as the reference line when the body stops.

The area occupied in the **acceleration–time graph** is a rectangle. The area of the rectangle is given by

A = l×b

From the above graph, the length of the rectangle is the acceleration, and breadth is the time; hence the equation is

A = a*t

But the area of the a-t graph is the velocity, then

v = a*t

v = 7× 8

v = 56 m/s.

Hence, by the definition of **the acceleration time graph, the area is nothing but the velocity.**

**How to find initial velocity with acceleration and time**?

When a body begins to travel from one point to another, initially, it possesses some velocity. A body doesn’t need constant velocity until it reaches its final destination. The velocity of the body changes with time as it travels through, and hence acceleration is acquired by the body.

**From the above explanation, it is clear that a traveling body may possess different velocities. The body’s velocity at the initial stage may differ from the final stage. Let us discuss finding velocity with acceleration and time at the initial point.**

Let us consider a car moving initially with the velocity v_{i}, and its velocity changes after a certain time t. The body is now accelerating with the acceleration ‘a’, and finally, when it reaches the endpoint, it has the velocity v_{f}.

The initial velocity can be calculated by three methods.

**By using the algebraic method:**

The acceleration due to change in the velocity is given by

a=(v_{f}-v_{i})/t

a*t = v_{f} – v_{i}

On rearranging

v_{i} = v_{f} – at

The above equation gives the initial velocity of the moving body.

**By the calculus:**

From the definition of acceleration, the equation is given by

a=dv/dt

Rearranging the terms;

adt = dv

Integrating the above equation by choosing the limits as the initial velocity vi at time t=0 and final velocity v_{f} at time t.

a ( t – 0) = (v_{f} – v_{i})

at = v_{f} – v_{i}

Rearranging the above equation, we get initial velocity.

v_{i} = v_{f} – at

**By the graphical method:**

A graph of velocity vs. time is plotted, whose slope gives the acceleration—then finding the slope, the initial velocity can be calculated.

From the above graph, we can say that.

**In a uniform time interval, the****velocity of the body changes.****OD is the time taken by the body to travel, and BD is the final velocity of the body.****A perpendicular lines BD to A is drawn parallel to OD. In the same way, a line BE is drawn parallel to OD.**

The above graph represents that,

The initial velocity of the body v_{i} = OA

The final velocity of the body v_{f} = BD

From the graph, BD = BC+ DC

Therefore, v_{f} = BC + DC

But DC= OA= v_{i}

v_{f} = BC + vi

From the graph, slope = acceleration a

a=BC/AC

But AC = t (from the graph)

a=BC/t

at = BC

Substituting the value of BC

v_{f} = at +v_{i}

v_{i} = v_{f} – at

**How to find change in velocity with acceleration and time**

In general, **the change in velocity with time gives acceleration.**

Let a body is moving with accelerating ‘a’ with time ‘t’ initially the object’s velocity is v_{i}, and at the final point, it has velocity v_{f}. Then change in the velocity is find out using the equation,

∆a=(Δv/Δt)

Where ∆v is the change in the velocity and ∆t is the change in time.

∆v= ∆a∆t

But the change in velocity is given by the **difference between the initial and final velocity.** It is given by the equation below.

∆v = v_{f} -v_{i}

**The change in the velocity can be calculated using the acceleration–time graph**. The area occupied under the a-t graph gives the change in the velocity.

Let us clearly understand it by considering the example represented by the graph given below.

The area covered in the acceleration time graph is a triangle. Hence, calculating **the change in velocity is given by calculating the area of the triangle.** The formula to find the area of the triangle is

A=(1/2)hb

Here, h is the height of the triangle, acceleration is considered the height, and b is the triangle’s base, which is given by the time axis. Thus the change in velocity is

∆v=(1/2)*6*9

∆v= 29 m/s.

Using the change in the velocity, we can find out the initial and final velocity of the body.

**Solved problems on how to find velocity with acceleration and time**

**Problem 1) A boat moves with an initial velocity of 11 m/s. The boat attains an acceleration of 3 m/s**^{2} for every 10 seconds. Then calculate the change in velocity and final velocity of the boat.

^{2}for every 10 seconds. Then calculate the change in velocity and final velocity of the boat.

**Solution:**

The data given for the calculation:

The initial velocity of the boat v_{i} = 11 m/s.

The change acceleration attain by the boat a = 3 m/s^{2}.

Change in time t = 10 sec.

∆v = ∆a∆t

∆v = 3 × 10

∆v = 30 m/s

To find the final velocity, the equation is

∆v = v_{f} -v_{i}

v_{f} = ∆v + v_{i}

v_{f} = 30 + 11

v_{f} = 41 m/s.

**Problem 2) The acceleration –time graph is given below. Find the change in the velocity and calculate the initial velocity if the final velocity is 54 m/s.**

**Solution:**

Given data:

Final velocity v_{f} = 54 m/s. From the acceleration –time graph, the area covered is the trapezium. So the area of the trapezium is given by,

A=[(a+b)/2)]*h

Where a, and b are the adjacent base of the trapezium, h is the height. From the graph; a = 9 units, b = 5 units, h = 4 units.

A=[(9+5)/2]*4

A = 28 units.

The change in velocity is equal to the area of the trapezium.

∆v = 28 m/s.

To find the initial velocity

∆v = v_{f} -v_{i}

v_{i} = v_{f} – ∆v

v_{i }= 54 – 28

v_{i} = 26 m/s.

**Problem 3) the acceleration –time graph is given to find the change in the velocity.**

**Solution:**

The above graph can be divided into three parts, represented in the dotted line, as shown in the figure below.

In the above graph, the following terms can be understood.

**OAD and BCE is the triangle; the area triangle is given by**

a=(1/2)hb

**ABCD is the rectangle; the area of the rectangle is given by**

A = l × b

In order to find the change in the velocity, the sum of the area of all the geometrical structures has to be calculated.

∆v = A=(1/2)hb+lb+(1/2)hb

Change in velocity ∆v = 180 m/s.

**Problem 4) Find the initial velocity of the ball, which is accelerating with 6m/s**^{2} with a time of 8 sec. The final velocity of the ball is 100 m/s.

^{2}with a time of 8 sec. The final velocity of the ball is 100 m/s.

**Solution:**

Data given: acceleration of the ball a = 6 m/s2.

Time t = 8 sec.

The final velocity v_{f} = 100 m/s.

To find the initial velocity of the body is given by the equation

v_{i} = v_{f} – at

v_{i} = 100 – (6 × 8)

v_{i} = 100 – 48

v_{i} = 52 m/s.

**Problem 5) Calculate the Change in the velocity of a moving object which has an initial velocity of 34 m/s. The acceleration of the object is 12 m/s**^{2}, and the change in time is 7 sec.

^{2}, and the change in time is 7 sec.

**Solution:**

Given:

Initial velocity of the object v_{i} = 34 m/s.

The acceleration of the object a = 12 m/s^{2}.

Change in time t = 7 sec.

The final velocity of the object is given by;

v_{f} = v_{i} + at

v_{f} = 34 + ( 12*7)

v_{f} = 34 + 84

v_{f} = 118 m/s.

The change in velocity is given by;

∆v = v_{f} – v_{i}

∆v = 118 – 34

∆v = 84 m/s.

**Problem 6) A disc moves with an initial velocity of 25 m/s. The disc changes its velocity every 10 sec. The change in the acceleration is 5 m/s**^{2}. Calculate the final velocity of the disc.

^{2}. Calculate the final velocity of the disc.

**Solution:**

Given data:

Initial velocity of the disc v_{i} = 25 m/s.

Change in acceleration ∆a = 5 m/s^{2}.

Change in time ∆t = 10 sec.

The change in velocity is

∆v = ∆a∆t

∆v = 5 × 10

∆v = 50 m/s.

The final velocity of the disc can be calculated using the formula given below

∆v = v_{f} – v_{i}

50 = v_{f} – 25

v_{f} = 50 + 25

v_{f} = 75 m/s.