In this article, we will see how to find velocity with acceleration, along with some examples, and will solve some problems.

**The acceleration of the object is directly proportional to the change in velocity with time. For an object accelerating in a circular motion or parabolic path, the velocity remains tangential to the arc.**

**How to Find Velocity from Angular Acceleration?**

Angular acceleration is defined as the change in angular velocity with respect to change in time duration and is represented as

a=Δω/Δt—(1)

**Angular velocity can be found out by calculating the variation in angle θ with respect to time. Hence,**

**because ω =d/dt—(2)**

**Therefore, we can write previous equation as**

therefore a=d^{2}θ/dt^{2}

From equation (1),

dω =adt

Integrating the equation

∫ dω =∫ adt

ω =at+C—(3)

When t=0, ω =ω_{0}

And hence, C=ω_{0}

Substituting this in the equation (3)

ω =ω_{0}+at —(4)

**This shows that the angular velocity of the object in a circular motion is equal to its initial angular velocity and the acceleration of the object into time.**

Consider a particle accelerating in a circular path with angular velocity ω

Let ‘s’ be the distance covered by the particle in time ‘t’. If the radius of the circular path is ‘r’ then ‘θ’ be the angle made by the particle displacing through a distance ‘s’.

Then, the linear velocity of the particle will be equal to the displacement of the particle in time ‘t’. The displacement here is ‘s’. Hence, the velocity is given as

v=Δs/Δt—(5)

The change in the angle ‘θ’ on the displacement of the particle is equal to the ratio of the arc length and the radius of the circle.

Δθ =s/r

therefore s=Δθr

Substituting this in the equation (5)

v=r Δθ/Δt

Since the angular velocity is equal to the change in angle with respect to time; we can rewrite the equation as

v=rω—(6)

Where ω is an angular velocity

This implies that **the linear velocity of the particle is the product of the radius of the circular path attained by the particle and the angular velocity**.

**Problem1:** A man standing in a gravitron of diameter=6m accelerates with a rate of 15 m/s. What must be the linear velocity of the gravitron. The initial velocity of the gravitron is 4 m/s. What is the acceleration of the gravitron at time 3min?

Given: Radius r=3m

Initial angular velocity ω_{0}=4m/s

Final angular velocity ω=15m/s

The linear velocity of the gravitron on attaining the angular speed of 15m/s.

v=rω

v=3*15=45m/s

The acceleration of the gravitron at a time of 3 min

ω=ω_{0}+at

15=4+a*3

11=a*3

a=11/3

a=3.67m/s^{2}

Hence, the acceleration at time t=3 min is 3.67m/s.

**Problem2:** A car accelerating in a circular track pick up the initial speed of 20km/h and accelerates at a speed of 15km/h^2. What is the velocity of the car after 15 minutes?

Given: ω_{0}=20 km/h

a=15 km/h^{2}

t=15 minutes=15/60=0.25 hr

Hence,

ω =ω_{0}+at

ω =20+15*0.25

ω =20+3.75=23.75 km/h

Therefore, the velocity of the car after 15 minutes will be 23.75 km/h.

**Relation between velocity, displacement, and Acceleration**

We have derived an equation on how to calculate the final velocity based on the acceleration and knowing the initial velocity of the system.

Considering the same eqn (4) from the above, we can write

v =v_{0}+at

Where v is a final velocity

v_{0} is a initial velocity

And a is an acceleration of the particle.

**The velocity is defined as the change in position of the object between the time interval.**

dx/dt=v_{0}+at

dx=(v_{0}+at)dt

Integrating the above equation

∫dx=∫(v_{0}+at)dt

]x=v_{0}t+1/2 at^{2} —(7)

**Since velocity is given by the displacement per unit time, displacement is equal to the product of average velocity and time.**

x=v^{→}t —(8)

Where v^{→} is a average velocity which is equal to v^{→}=v_{0}t+v/2

From eqn (4) we gets t=v-v_{0}/a

Substituting this in the above eqn(), we have

x=v+v_{0}/2*v-v_{0}/a

x=v^{2} -v_{0}^{2}/2a —(9)

Rearranging this equation

v^{2}=v_{0}^{2}+2ax—(10)

This is another kinematic equation for a particle in a rectilinear motion.

**How to Find Velocity from Centripetal Acceleration?**

The velocity of the object accelerating in a circular path is perpendicular to the direction of the centripetal force exerting inward.

**The centripetal force and velocity of the object in motion is given by the relation**

** Fc=mv ^{2}/r—(11)**

**Where r is the radius of the circle**

**V is a linear velocity**

**M is a mass of the particle**

An object of mass ‘m’ accelerating in a circular path of radius ‘r’, the linear velocity is equal to the radius of the circular path and the angular velocity of the particle

v=rω

Where ω is the angular velocity of the particle

And, force is equal to the product of mass into the acceleration of the object.

Substituting this in the equation (7);

F=mr^{2}ω^{2}/r

F=mrω^{2}

ma=mrω^{2}

a=rω^{2} —(12)

Hence, the acceleration and the velocity of the particle in a centripetal motion are related by the equation (8) according to which **the acceleration of the particle in a motion is a product of the radius of the circular path and the square of the angular velocity attained by the particle**.

**Problem3:** A boy tied a stone at one end of the rope of length 1m and holds another end of the rope in his hand and rotates in a circular motion at a rate of 2 rotations per second. Calculate the angular velocity of the stone?

Solution: Since the length of the rope is 1m, the radius of the circular path is equal to 1m.

In 1 second a stone elapse 2 rotations which are equal to the two circumferences of the circular path elapsed by the stone.

The circumference of the circular path

C=2π r=2π* 1=2π

Hence, the stones covers a distance of 2*2π =4π in one second.

Therefore the angular acceleration of the stone is

a =4π /s

Hence the angular velocity of the stone is

therefore a=rω^{2}

4π =1*ω^{2}

ω =√4π =0.6m/s

**Problem4:** A ball of radius 0.3m traveling with a velocity of 5m/s in a circle of diameter 5m. What is the angular velocity of the ball?

Given: r=5m

V=5m/s

Using equation v=ωr

The angular velocity of the ball is

ω=v/r

ω=5/5=1rps

**Read more on centripetal acceleration.**

**How to Find Velocity from Variable Acceleration?**

An object is said to be moving with variable acceleration if its velocity changes frequently in different time intervals.

**If the acceleration of the particle is ‘a’, then a=dv/dt which is variable with time ‘t’. The velocity can be calculated by integrating the equation dv=adt.**

Consider a particle accelerating with velocity v1 in a random motion. If the particle suddenly changes its direction and the velocity from v_{1} to v_{2} after time interval t_{1} to t_{2}.Then the acceleration a_{1} of the particle is

a_{1}=v_{2}-v_{1}/t_{2}-t_{1}

If at time t_{1}=0, v_{1}=0, and at t_{2}=30 seconds, v_{2}=3m/s, then,

a_{1}=3-0/30-0=3/30=0.1m/s^{2}

Again, the particle change the direction and attains the velocity v_{3} at time t_{3}.

Now the acceleration due to change in the velocity of the particle becomes

a_{1}=v_{3}-v_{3}/t_{3}-t_{3}

If at t_{3}=60 seconds v_{3} =8m/s,

a_{1}=8-3/60-30=5/30=0.167m/s^{2}

Hence change in acceleration now due to random motion of the particle is

Δa=a_{2}-a_{1}=0.167-0.1=0.067 m/s^{2}

Which is approximately0.07m/s^{2}

**Problem5:** If the acceleration of the particle is given by the equation a=6t^{2}+4t, find the velocity of the particle at time t=2 seconds.

Solution: a=6t^{2}+4t

dv/dt=6t^{2}+4t

dv=(6t^{2}+4t)dt—————(13)

The above equation is variable with time ‘t’, hence it is called the variable acceleration because time is not constant.

Integrating equation (13)

∫dv=∫6t^{2}+4dt

v=6t^{3}/3+4t^{2}/2

v=2t^{3}+2t^{2}

v=2(t^{3}+t^{2})

When time t= 2 seconds

v=2(2^{3}+2^{2})

v=2(8+4)

v=2*12=24 m/s

Therefore the velocity of the particle is 24m/s.

**How to Find Velocity with Acceleration and Radius?**

When an object accelerates in a circle, it exerts a centripetal force directing towards the center of the circle.

**If ‘r’ is the radius of the circle, and ‘m’ is the mass of the object, then the centripetal force exerted on the object is given by**

F_{c}=mv^{2}/r

Since, F_{c}=ma

ma=mv^{2}/r

a=v^{2}/r—(14)

v=√ar—(15)

**Hence, velocity is directly proportional to the square root of the product of acceleration and radius of the circle.**

**How to Find Velocity with Acceleration and Angle?**

The acceleration is defined as the ratio of change in angular velocity with respect to time.

**For an object moving in a circular path, the velocity and hence the acceleration of the object is measured in terms of change in angle θ.**

a=dω/dθ

Using the above eqn (4)

ω =ω_{0}+at

Since, ω =dθ /dt

Hence,

dθ /dt=ω_{0}+at

dθ=( ω_{0}+at ) dt

Integrating this equation

∫dθ=∫(ω_{0}+at ) dt

θ=ω_{0}+1/2 at^{2}

ω_{0}=θ t^{-1}-1/2 at —(16)

The above equation shows the relation between velocity \omega _0, acceleration ‘a’ and angle θ.

**Problem6:** The angular speed of the motor is increased from 1800rpm to 2400 rpm in 10 seconds. Find the angular acceleration and number of revolutions of the motor during this time?

The initial angular velocity in rad/sec

ω_{0}=2π*1800

=2π*1800/60=60π rad/s

The final angular velocity in rad/sec

ω =2π*2400

=2π*2400/60=80πrad/s

Angular acceleration a=ω-ω_{0}/t

a=(80-60)π/10=2π rad/s^{2}

The angular acceleration of the motor is 2π rad/s^{2}

Angular displacement in time ‘t’ is given by

θ=ω_{0}t+1/2 at^{2}

=60π*10+1/2 * 2 Pi π *10^{2}

=600π +100π=700π

Number of revolution = 700π/2π=350

Hence, the motor takes 350 revolutions per second.

**How to Find Velocity with Acceleration and Force?**

The normal force is defined as the product of mass and acceleration whereas the force applied on the object is equal to the ratio of the work done and the displacement of the object.

**In centripetal motion, the force is proportional to the square of the velocity of the object tracing the circular path and the mass of the object and inversely proportional to the separation of the object from the center of the circular path.**

In linear motion, the final velocity of the object is related to the acceleration by the equation

v=u+at

Since, F=ma

a=F/m

Substituting this in the above equation

v=u+F/mt

When the object is in a circular motion, velocity is related to the acceleration by the relation

a=v^{2}/r

Hence, the velocity is related to the force by the equation

v^{2}=Fr/m

**Read more on How To Find Final Velocity Without Acceleration: Facts, Problems, Examples.**

**Frequently Asked Questions**

**How acceleration is dependent on time and velocity?**

The acceleration is variable with time and equals the change in velocity of the object with time.

**The acceleration is depends on the time and the velocity of the object by the relation **

**v=u+at. Hence, a=v-u/t**

**What is the difference between speed and velocity?**

Speed is a scalar quantity whereas velocity is a vector quantity.

**Speed is measured in terms of the path traveled by the object in time ‘t’ whereas velocity is not concerned about the path covered by the object but the initial and final position of the object.**

**Why do we experience sudden jerk backward on accelerating the car?**

On acceleration, the velocity of the car in motion changes.

**A change in velocity changes the momentum of the car at the same time and experiences a force that is felt on the body. This can be represented by the relation as** F**=ma=mdv/dt=d/dt(mv)=dp/dt**