Skip to content
Related Articles

Related Articles

Improve Article

Uniformly Accelerated Motion

  • Last Updated : 14 Jul, 2021

When a body moves in a plane or on a straight line, three parameters are used to describe its motion – distance, velocity, and acceleration. Distance or displacement is self-explanatory. Velocity represents the rate of change of position, while acceleration represents the rate of change of velocity. All three quantities are vector quantities. Acceleration can be uniform or non-uniform. A uniform acceleration has a constant value and direction. It is essential to know the equation of motions that describe the motion of an object under uniform acceleration. Let’s look at them in detail. 

Acceleration

Acceleration is defined as the rate of change of the velocity vector. Acceleration can be constant or varying. In the case of constant acceleration, its value is given by the ratio of net change of velocity and the total time taken. It is also called average acceleration. In the cases where the acceleration varies with time, instantaneous acceleration is calculated. 

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Course, specifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

Average Acceleration: 



\vec{a} = \frac{\vec{v_f} - \vec{v_i}}{\Delta t}

Instantaneous acceleration: 

\vec{a} = \frac{d\vec{v}}{dt} \\ = \vec{a} = \frac{dx}{dx}\hat{i} + \frac{dy}{dt}\hat{j}

Uniform Acceleration

Uniform acceleration is the acceleration that does not vary with time. In such cases, the rate of change of velocity remains constant. Since acceleration is a vector quantity, even the direction of motion remains the same in the case of constant acceleration. Since the body is moving in a single direction with a constant magnitude of acceleration, vector notations can be dropped. 

Some examples of constant acceleration include 

  1. Free-falling object.
  2. A ball rolling down a frictionless slope.
  3. A bicycle whose brakes have been engaged.

Equations of Uniformly Accelerated Motion

In this case, keeping the value of acceleration constant. Equations of motion can be described. Let’s say the initial velocity of an object was “u”, now a constant force is applied which causes the body to move with constant acceleration “a” and the body reaches the velocity in time “t” while covering the distance “s”.

First Equation of Motion

In the case of constant acceleration, its value is given by, 



a = \frac{v - u}{t} \\ =at = v - u \\ =v = u + at

Second Equation of Motion 

Instantaneous velocity is given by, 

v = \frac{ds}{dt}

This equation can be re-arranged in the following form,

ds = vdt  

Substituting the value of velocity from the previous equation, 

ds = (u + at)dt 

Integrating both sides, 

\int^{s_2}_{0}ds = \int^{t}_{0}(u + at)dt \\ s = \int^{t}_{0}udt + \int^{t}_{0}atdt \\ s = u[t]^{t}_{0} + a[\frac{t^2}{2}]^{t}_{0} \\ s= ut + \frac{1}{2}at^2



Third Equation of Motion

Instantaneous acceleration and instantaneous velocity is given by, 

a = \frac{dv}{dt}

v = \frac{ds}{dt}

Cross multiplying both of these equations, 

a\frac{ds}{dt} = v\frac{dv}{dt} \\ = \int^{s}_{0}a\frac{ds}{dt} = \int^{v}_{u}v\frac{dv}{dt} \\ = as = [\frac{v^2}{2}]^{v}_{u} \\ = as = \frac{v^2 - u^2}{2} \\ = 2as =v^2 - u^2

v2 = u2 + 2as

Sample Problems

Question 1: If a body is moving at an acceleration of 2 m/s2. If the initial speed was 15m/s, what will be the speed in 5 seconds. 

Answer: 

Let u denote the initial velocity and v denote the final velocity. 



Given: u = 15m/s, a = 2 m/s2 and t = 5 

For finding out the value of “v”, first equation of motion can be used. 

v = u + at 

Plugging the values in this equation, 

v = u + at 

⇒ v = 15 + (2)(5) 

⇒ v = 15 + 10 

⇒ v = 25 m/s 

Question 2: If a body is moving at an acceleration of -5 m/s2. If the initial speed was 30m/s, what will be the distance co in 5 seconds. 

Answer: 

Let u denote the initial velocity and v denote the final velocity. 

Given: u = 40m/s, a = -5 m/s2 and t = 5 

For finding out the value of “v”, the first equation of motion can be used. 

v = u + at 

Plugging the values in this equation, 

v = u + at 

⇒ v = 30 – (5)(5) 

⇒ v = 30 – 25 

⇒ v = 5 m/s 

Question 3: If a body is moving at an acceleration of -5 m/s2. If the initial speed was 40m/s, what will be the speed in 5 seconds. 

Answer: 

Let u denote the initial velocity 

Given: u = 40m/s, a = -5 m/s2 and t = 5 

For finding out the value of “s”, the first equation of motion can be used. 

s= ut + \frac{1}{2}at^2

Plugging the values in this equation, 

s= ut + \frac{1}{2}at^2 \\ = s = (40)(5) + \frac{1}{2}(-5)(5)^2 \\ = s = 200 + \frac{1}{2}(-125) \\ = s = \frac{400 - 125}{2} \\ = s=  \frac{275}{2} \text{ m}

Question 4: If a body is moving at an acceleration of 10 m/s2. If the initial speed was 20m/s, what will be the speed in 2 seconds. 

Answer: 

Let u denote the initial velocity 



Given: u = 20m/s, a = 10 m/s2 and t = 2 

For finding out the value of “s”, the first equation of motion can be used. 

s= ut + \frac{1}{2}at^2

Plugging the values in this equation, 

s= ut + \frac{1}{2}at^2 \\ = s = (20)(2) + \frac{1}{2}(10)(2)^2 \\ = s = 40 + 20 \\ = s=  60 \text{ m}

Question 5: A racing car catches a speed of 20m/s in 2 seconds. Find the distance covered by the car in the process. 

Answer: 

Let u denote the initial velocity and v denote the final velocity. 

Given: u = 0 m/s, v = 20m/s and t = 2. 

For finding out the value of “a”, the first equation of motion can be used. 

v = u + at 

Plugging the values in this equation, 

v = u + at 

⇒ 20 =0 + (a)(2) 

⇒ 20 = 2a 

⇒ a = 10 m/s2

 For finding out the distance, a third equation of motion will be used. 

v2 = u2 + 2as

⇒ 202 = 0 + 2(10)s

⇒ 400 = 20s 



⇒20m = s

Question 6: A rocket catches a speed of 50m/s in  5seconds. Find the distance covered by the rocket in the process. 

Answer: 

Let u denote the initial velocity and v denote the final velocity. 

Given: u = 0 m/s, v = 50m/s and t = 5. 

For finding out the value of “a”, the first equation of motion can be used. 

v = u + at 

Plugging the values in this equation, 

v = u + at 

⇒ 50 =0 + (a)(5) 

⇒ 50 = 5a 

⇒ a = 10 m/s2

 For finding out the distance, a third equation of motion will be used. 

v2 = u2 + 2as

⇒ 502 = 0 + 2(10)s

⇒ 2500 = 20s 

⇒125m = s




My Personal Notes arrow_drop_up
Recommended Articles
Page :