How To Find Velocity Without Time: Facts, Problems, Examples


In this article, we are going to learn various ways on how to find velocity without time along with examples, some facts, and how to solve related problems.

The energy of the object is conserved, based on this fact; the velocity of the object is equal to the square root of twice the product of its acceleration and the distance that it elapses, also depending upon the initial velocity of the object.

How to Find Final Velocity withoutTime?

The accelerating object changes its velocity with time.

The velocity attained by the object over the period of time until it stops accelerating for the duration is said to be the final velocity of the object.

Let us see how to find the final velocity without using the time character.

Consider a velocity-time graph showing the variation in the velocity of the object in a uniform linear motion with respect to time. From the graph, we can read that the time T=0, the velocity=u, and at time T=t, the velocity=v.

how to find final velocity without time
Velocity-Time Graph

Since velocity is the ratio of change in position with varying time, displacement will be equal to

x=vt —(1)

The above graph is related to the displacement by the relation as shown in the eq(1).

Let us measure the area covered by the object, the total area will be equal to the sum of the area of the triangle (∆ABC) and the quadrilateral (□ACDO).

x = Ar(∆ ABC)+ Ar(∆ ACDO)

=1/2 bh+lb

=1/2 t*(v-u)+ut—(3)

Since acceleration is equal to the change in velocity with time, i.e.

a=dv/dt —(4)

a=v-u/t-0=v-u/t

v-u=at—(5)

Substituting eq(5) in the equation (3)

x=1/2 t * at+ut

x=1/2 at2+ut—(6)

Also,

From equation (4), we have

dv=adt

Integrating this equation we get

∫dv=∫dt

v=at+C

At t=0, v=u, hence, C=u

Therefore,

v=u+at —(7)

Now, this equation is a time dependent equation, and time ‘t’ from the above relation is equal to

t=v-u/a —(8)

The average velocity is the sum of all the velocities attained by the object in different time intervals divided by the total number of velocities summed together. Here, we have two velocities, the initial velocity ‘u’ and the final velocity ‘v’, therefore the average velocity is

Vavg=Vfinal+Vinitial/Total Number of velocities

Vavg=(v+u)/2 —(9)

Using eqn (1), x=vt

Substituting eqn (8) & (9) in eqn (1)

x=(v+u)/2 *(v-u)/a

x=v2-u2/2a

2ax=v2-u2/2

v2=u2+2ax —(10)

The above equation is independent of time and shows the relation between the initial velocity of the object, the constant acceleration, and the displacement of the object.

Problem1: A ball is moving in a rectilinear motion with an acceleration of 2m/s. If the initial velocity of the ball is 4m/s, what will be its velocity when it will cover a distance of 20 meters?

Given: a=2m/s

u= 4m/s

d=20m

Using equation (10),

v2=u2+2ax

=42+2*2*20

=16+80=96m/s

therefore v=9.8m/s

Hence, when the ball will cover a distance of 60 meters, the velocity of the ball will be 9.8 m/s.

How to Find Velocity of a Falling Object without Time?

Linear velocity is time-dependent and is a ratio of change in position along with time.

An object falling is accompanied by energies within it, in the form of kinetic energy and potential energy, and energy can neither be created nor can it be vanished. Based on this fact we can calculate the velocity of the object irrespective of time.

When the object is raised at a height from above the ground it gains some potential energy which is then converted into kinetic energy and is utilized during its flight.

Consider an object of mass ‘m’ kept on the table of a height h1, it experiences an external force and gains momentum, and starts accelerating towards the ground. Since the object is at rest on the table its initial velocity u = 0 and hence the kinetic energy is also nil. The object when it is at height h1 has the potential energy U1 associated with it.

U1=mgh1

While commencing its journey towards the ground, this potential energy is converted into the kinetic energy

K.E2=1/2mv2

After falling on the ground the potential energy of the object U2=mgh0; since h0=0, U_2=0.

Since the energy of the object is conserved, the sum of the kinetic energy and potential energy before and after falling on the ground will equal.

K.E1+U1=K.E2+U2

U1=K.E2

mgh1=1/2mv2

v2=2gh1

v=√2gh1—(11)

Hence the velocity of the object falling towards the ground due to gravity is given by the equation (11).

Problem2: A boy is playing with a ball. He threw the ball high up in the air and observes the free fall of the ball. What will be the velocity of the ball while approaching down to the ground if the ball raises at a height of 8 meters above the surface of the Earth?

Given: Height h=8m,

g= 9.8m/s2

Using equation (11),

v=√2gh1

=√2*9.8*8

=√156.8=12.52m/s

Hence, the final velocity of a ball approaching the ground will be equal to 12.5m/s.

How to Find Horizontal Velocity without Time?

An object moving in a horizontal motion irrespective of the acceleration due to gravity of the Earth and the applied force then is said to be horizontal velocity.

Horizontal velocity in simplicity is equal to the ratio of the distance traveled by the object and the time taken to cover the distance. That is,

Horizontal velocity VH=Distance travelled/Time taken

For an object traveling in a projectile motion, the object is associated with two velocity components, the horizontal component in x-axis ‘V Cosθ’ in the direction of the motion, and a vertical component in y-axis ‘V Sinθ’ acting upward while accelerating in the upward direction and then downward in negative y-axis while accelerating towards the ground.

how to find velocity without time
Graph of projectile motion showing the constant horizontal velocity

From the above graph, to calculate the horizontal velocity which is constant and in the direction of the x-axis, the Cosine component by trigonometry is

Cosθ=adjacent/hypotenuse =Horizontal Velocity/Initial Velocity

Cosθ=VH/V

VH=V Cosθ —(12)

The above relation shows the equation to find out the horizontal velocity independent of time.

Example: A ball is thrown in the air traveling in a parabolic path making an angle of 600 with the surface of the Earth. If the initial velocity of the ball is 5 m/s, find the horizontal velocity of the ball.

Given: θ=600

Initial velocity u=5m/s

Using the equation,

VH=VCosθ

=5*Cos(60)

=5*1/2=2.5m/s

Therefore the horizontal velocity of the ball is 2.5 m/s.

The range of a projectile is how much distance an object will cover from its initial point that is at point (0,0) in the above graph depending upon the horizontal velocity of the object and for how long the object is in the air.

That is,

R=VHTf—(13)

Where R is a range, VH is the horizontal velocity of the object and Tf is a time of flight.

Time taken by the object during its projectile motion to return back on the ground at y=0 is mentioned as the time of flight.

Let us derive an equation for a time of flight using the equation of rectilinear motion given below

V=U+at—(14)

The initial velocity of the object is U=VSinθ

The final velocity V Cosθ =0

And a=-g since the acceleration is in the negative y-axis.

The equation becomes,

V= V Sinθ –gt

Since final velocity is equal to zero,

0= VSinθ –gt

V Sinθ =gt

t=V Sinθ/g —(15)

This is the time required for an object to attain the maximum height during the flight.

This implies that the time for maximum height will be equal to the time required for an object to cover the rest half of the flight.

Hence, the time for a flight

Tf=2 V Sinθ/g —(16)

Substituting eqn (12) &eqn (16) in the equation (13),

R=V\ Cosθ*2V Sinθ/g

R=V2/g* 2SinθCosθ

R=V2 Sin2θ/g —(17)

Hence, velocity of the object in projectile motion is also equals to

V=√Rg/Sin2θ —(18)

The velocity can be calculated on measuring the range of flight and the angle that the object makes with the ground.

Read more on Projectile Motion.

How to Find Centripetal Velocity without Time?

An object moving in a circular path along with time acquires centripetal velocity.

The direction of the velocity of the object in a circular path remains tangential to the circle and perpendicular to the centripetal force directing towards the center.

Consider an object of mass ‘m’ accelerating along a circular path due to the external force applied on the object. The centripetal force acting on the object is directly proportional to the square times the velocity attained by the object and inversely proportional to the distance of the object from the center of the circle. The force applied is equal to the centripetal force experienced on the object.

F=Fc

ma=mv2/r

a=v2/r

v2=ar

v=√ar—(19)

The velocity of the object in a circular motion is equal to the square root of the acceleration of the object and the radius of the circular track and is independent of the time.

Example: Consider a car traveling in a circular pathway from outside the football playground with an acceleration of 40km/h. The diameter of the ground is 80 meters. Find the velocity of the car.

Given: a=40km h=40*1000/60*60=11.1m/s

d=80m, r=80/2=40m

v=√ar

=√11.1m/s*40m

=√444

=21.1m/s2

=75.96 km/h~ 76 km/h

Hence, the velocity of the car accelerating in a circular path is 76km/h.

Read more on How To Find Velocity With Acceleration: Different Approaches, Problems, Examples.

Frequently Asked Questions

Q1. Two girls are playing passing the ball; one girl throw the ball high in the air making an angle 450 with the direction of the horizontal velocity passing the ball to the girl standing at a distance of 10 m away from her. What is the velocity attained by the ball on its throw?

Given: θ=450

Range of the flight of ball on throw R=10 meters

R=V2 sin2θg

V=√Rg sin2θ

V=√10* 9.8/Sin(2*60)

V=√98/Sin(120)

V=√98/0.86

V=√113.95

V=10.67 m/s

Hence the velocity of the ball during its flight is 10.67 m/s.

What is the average velocity?

The accelerating object changes its direction of velocity and speed along with a certain duration of time.

The sum of all the velocities varied with time divided by the total number of variations is called the average velocity.

AKSHITA MAPARI

Hi, I’m Akshita Mapari. I have done M.Sc. in Physics. I have worked on projects like Numerical modeling of winds and waves during cyclone, Physics of toys and mechanized thrill machines in amusement park based on Classical Mechanics. I have pursued a course on Arduino and have accomplished some mini projects on Arduino UNO. I always like to explore new zones in the field of science. I personally believe that learning is more enthusiastic when learnt with creativity. Apart from this, I like to read, travel, strumming on guitar, identifying rocks and strata, photography and playing chess. Connect me on LinkedIn - linkedin.com/in/akshita-mapari-b38a68122

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