# How To Find Velocity With Constant Acceleration: Problems And Examples

The particle’s motion under constant acceleration gives much useful stuff regarding the entities that describe its motion. Under constant acceleration, the velocity variation with time can easily be monitored.

Constant refers to the steadiness of the system. The steadiness in the acceleration means the velocity is changing at a steady rate; it does not mean that velocity is constant. The variation of velocity with constant acceleration leads to the question of how to find velocity with constant acceleration?

Before going to find the velocity with constant acceleration, we need certain expressions to be remembered, which are listed below that are essential to describe the particle as accelerated.

• The change in the position of the particle gives the distance traveled by the particle; it can be expressed as ∆x = x-x0
• The change in the velocity of the particle is ∆v = v-v0
• The change in the time interval ∆t = t-t0

Under constant acceleration, the velocity can be described as the average of the initial and final velocity of the moving particle. So to find velocity with constant acceleration, we need to assume that the instantaneous acceleration and average acceleration are equal to the constant acceleration; these assumptions avoid the use of the calculus method for finding the velocity.

The average velocity of a particle moving with an initial velocity of v0 and final velocity v is given by

va=v+v0/2

The acceleration of the particle is given by

a=Δv/Δt

Since we have assumed that the particle is traveling with an initial velocity, thus the initial time is considered to be zero, and the time at velocity v is t. Thus the equation of acceleration can be rewritten as

a=v-v0/t0

a=v-v0/t

Thus rearranging the above equation, we get

v-v0 = at

At constant acceleration, we take the average of the initial and final velocity, thus adding v0 and dividing by 2 on both the side we get,

v-v0 +v0/ 2 =(v0 + at) / 2

va – v0/ 2=(v0 + at) / 2

Thus we get equation of velocity as

va=v0+at/2

If the initial velocity is zero, then the average velocity is equal to the final velocity of the moving body; thus, we can write the expression as

v=at/2

This gives the expression for velocity with constant acceleration.

## Finding initial velocity with constant acceleration, distance, and time

Suppose a particle is moving with an initial velocity of v0 m/s, and it begins to accelerate for every t seconds at a constant rate; then, how to find the velocity with constant acceleration?

The initial velocity is measured when the particle just begins its motion. At that instance, the time is zero, and the particle’s position is also at the origin. After a certain time, if the particle begins to accelerate at a constant rate and covers the distance x, the initial velocity can be calculated as follows:

The distance traveled by the particle from o to x is given by

x = vt

Since we are dealing with the constant acceleration problem, we consider the average velocity for the calculation.

x = vat

Substituting the value of average velocity in a given equation, we get the distance traveled by the particle as

x=v+v0/2t

Since we have unaware of the final velocity attained by the particle, however, we know from the kinematic equations

v= v0 +at

Substituting the value of v in the distance equation,

x=v0 +at + v0 /2t

x=2v0 +at2 /2

Rearranging them, we get

v0=x/t – 1/2 at

The above equation gives the initial velocity constant acceleration, distance, and time.

## Finding final velocity from the constant acceleration and distance

The constant acceleration depends on the final velocity of the moving particle; however, this dependency of final velocity on the acceleration lets us know about the stuff like how significant the acceleration is and how steady the velocity is.

Any particle whose final velocity is given means an initial velocity exists at the beginning of the motion. The final velocity will be equal to the initial velocity only when the particle is not accelerating.

Considering the facts mentioned above the acceleration can be written as

a=v-v0

The distance traveled by the particle with time gives the velocity. If the particle travels the distance from position x0 to x, at time t=0 to t=t, the velocity is

v=v-v0/t

From the above equation, the time factor can be expressed as

t=v-v0/v

Substituting the value of t in the acceleration, we get

a=(v-v0)/(x-x0)

Since we are talking about the constant acceleration, the value of velocity is determined as the average velocity, thus the above equation becomes,

a=[(v-v0)/(x-x0)]/ (v+v0)

Rearranging the terms, we get

a=(v-v0/v+v0)2(x-x_0)

2a(x-x0) = v2 – v02

v2 = v02+2a(x-x0)

This gives the final velocity of the particle at a=constant.

## Problem 1) In a running race, a runner runs with a certain velocity. He is accelerating at a constant rate of 4 m/s2 for every 5 seconds. He covers a distance of 55 m. Find initial velocity with constant acceleration, distance, and time are given.

Solution:

Data required for the calculation:

Acceleration of the runner is a = 4 m/s2.

Time taken to accelerate t = 5 seconds.

Distance covered by the runner x = 55 m.

Strategy:

From the given data, the initial velocity of the runner is given as

v0=x/t-1/2 at

v0=55/5-1/2 (4*5)

v0 = 11-10

v0 = 1 m/s.

## Problem 2) A ball is rolling on a solid ramp with an initial velocity of 6 m/s. As the ramp is slightly inclined, it begins to accelerate with a constant rate of 2 m/s2. The initial velocity attained by the ball is at a distance of 3 meters, and it covers a distance of 27 meters. Calculate the final velocity of the ball and hence find the change in the velocity.

Solution:

Data given:

Initial velocity of the ball v0 = 6 m/s.

The initial position of the ball x0 = 3m.

The final position of the ball x = 27m.

The constant acceleration of the ball a = 2 m/s2.

Strategy:

The final velocity of the rolling ball is given by the expression

v2 = v02+2a(x-x0)

v2 = 6+2(2)(27-3)

v2 = 6+96

v2 = 102

v=√102

v = 10.09 m/s.

## Problem 3) How to find the velocity of an object which is accelerating with a constant rate of 6 m/s2 for every 13 seconds?

Solution:

Given – The constant acceleration of the moving object a = 6m/s2.

The time taken by the object to accelerate t = 13s.

Strategy:

The problem is given with only constant acceleration and time, since the object is accelerating in a constant rate, its velocity can be expressed as

v=at/2

v=6*13/2

v = 39m/s.

## Problem 4) The final velocity of a car is given by 14 m/s. if it accelerates for every 9 seconds by 4 m/s2, calculate the distance traveled by car and hence find the initial velocity

Solution:

The data provided for the calculation:

Final velocity of the car v = 14m/s.

The acceleration of the car a = 4m/s2.

Time taken by the car to accelerate t = 9s.

Since we have only final velocity, acceleration and time, so let us assume the car is initial moving with velocity zero. Thus the equation is given by

v2 = 0 +2a(x-x0)

(14)2 = 2(4) (x-x0)

196 = 8(x-x0)

x-x0 = 24.5m

The obtained expression it is clear that the distance traveled by the car is 24.5 meters.

The initial velocity is given by

v0=x/t – 1/2 at

v0=24.5/5-1/2(4*9)

v0 = -13.1 m/s.

The negative sign indicates that the car is moving in the opposite direction of the intended direction.

## Problem 5) an aircraft is moving with an initial velocity of 76m/s on a runway and travels a distance of 54 m before it can land. The constant velocity of the aircraft is 8 m/s2. Calculate the final velocity and time taken by the aircraft to cover the distance.

Solution:

Data are given for the calculation:

Initial velocity of the aircraft v0 = 76m/s.

Acceleration of the aircraft a = 8m/s2.

Distance traveled by the aircraft x = 34m.

The final velocity is given by

v2 = v02+2a(x-x0)

v2 =76+2(8)(34-0)

v2 = 76+544

v2 = 620

v = 24.89 m/s.

The time taken by the aircraft

t=v-v0/a

Since the initial velocity is more than the final velocity, the aircraft is decelerating; thus, we can write the equation as

t=v0-v/a

t=34-24.89/8

t=1.13 seconds.

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