Physics Problems with Solutions

  • Uniform Acceleration Motion: Problems with Solutions

Problems on velocity and uniform acceleration are presented along with detailed solutions and tutorials can also be found in this website.

From rest, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 1

With an initial velocity of 20 km/h, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 2

A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds. a) What is the acceleration of the car in m/s 2 ? b) What is the position of the car by the time it reaches the velocity of 72 km/h? Solution to Problem 3

An object is thrown straight down from the top of a building at a speed of 20 m/s. It hits the ground with a speed of 40 m/s. a) How high is the building? b) How long was the object in the air? Solution to Problem 4

A train brakes from 40 m/s to a stop over a distance of 100 m. a) What is the acceleration of the train? b) How much time does it take the train to stop? Solution to Problem 5

A boy on a bicycle increases his velocity from 5 m/s to 20 m/s in 10 seconds. a) What is the acceleration of the bicycle? b) What distance was covered by the bicycle during the 10 seconds? Solution to Problem 6

a) How long does it take an airplane to take off if it needs to reach a speed on the ground of 350 km/h over a distance of 600 meters (assume the plane starts from rest)? b) What is the acceleration of the airplane over the 600 meters? Solution to Problem 7

Starting from a distance of 20 meters to the left of the origin and at a velocity of 10 m/s, an object accelerates to the right of the origin for 5 seconds at 4 m/s 2 . What is the position of the object at the end of the 5 seconds of acceleration? Solution to Problem 8

What is the smallest distance, in meters, needed for an airplane touching the runway with a velocity of 360 km/h and an acceleration of -10 m/s 2 to come to rest? Solution to Problem 9

Problem 10:

To approximate the height of a water well, Martha and John drop a heavy rock into the well. 8 seconds after the rock is dropped, they hear a splash caused by the impact of the rock on the water. What is the height of the well. (Speed of sound in air is 340 m/s). Solution to Problem 10

Problem 11:

A rock is thrown straight up and reaches a height of 10 m. a) How long was the rock in the air? b) What is the initial velocity of the rock? Solution to Problem 11

Problem 12:

A car accelerates from rest at 1.0 m/s 2 for 20.0 seconds along a straight road . It then moves at a constant speed for half an hour. It then decelerates uniformly to a stop in 30.0 s. Find the total distance covered by the car. Solution to Problem 12

More References and links

  • Velocity and Speed: Tutorials with Examples
  • Velocity and Speed: Problems with Solutions
  • Acceleration: Tutorials with Examples
  • Uniform Acceleration Motion: Equations with Explanations

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problem solving uniformly accelerated motion

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Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

  • (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

physicsgoeasy

Uniform Acceleration

PhysicsGoeasy

  • September 13, 2023
  • Kinematics , Mechanics

Uniform acceleration is a specific type of motion in which an object’s velocity changes at a constant rate over time. It serves as a foundational concept in physics, particularly in the study of mechanics. The purpose of this article is to give readers a thorough understanding of uniform acceleration by looking into its definition, characteristics, kinematic equations, graphical representations, and real-world examples.

Table of Contents

Definition of Uniform Acceleration

If the velocity of a body changes by an equal amount in an equal interval of time, however, small the interval may be then its acceleration is said to be uniform.

Uniform acceleration (\( a \)) can also be defined as the rate of change of velocity per unit time, where this rate of change remains constant. In mathematical terms, uniform acceleration is defined by the equation:

\[ a = \frac{{\Delta v}}{{\Delta t}} \]

Here, \( \Delta v \) represents the change in velocity, and \( \Delta t \) denotes the change in time.

Characteristics of Uniform Acceleration

Understanding the nuances of uniform acceleration becomes easier when we delve into its core characteristics. Here is a detailed breakdown:

Constant Rate of Change in Velocity

In uniform acceleration, the velocity of an object changes at a consistent rate.

Applicability of Kinematic Equations

Uniformly accelerated motion can be described using three fundamental kinematic equations: 1. \( v = u + at \) 2. \( s = ut + \frac{1}{2}at^2 \) 3. \( v^2 = u^2 + 2as \)

Linear Velocity-Time Graph

A velocity-time graph for uniform acceleration is a straight line, and its slope equals the value of the constant acceleration.

Parabolic Displacement-Time Graph

The displacement-time graph for such motion is a parabola, arising from the quadratic relationship $(\frac{1}{2}at^2)$ between displacement and time in the second equation of motion.

The figure given below shows the distance-time graph of an object moving with a uniform accelerated motion where acceleration $a=2m/s^2$.

distance time graph

Importance of Initial Conditions

Initial velocity (\( u \)) and initial position (\( s_0 \)) are essential parameters for problem-solving in uniformly accelerated motion. These conditions act as the starting point for any calculations involving kinematic equations.

Idealized Conditions

It is often an idealized model but serves as a practical approximation for real-world applications like free-falling objects in a vacuum or a car accelerating on a straight road.

Directional Aspects

Both the velocity and acceleration vectors can be in the same or opposite directions, determining whether the object speeds up or slows down.

Kinematic Equations for Uniformly Accelerated Motion

Uniformly accelerated motion can be modeled using the following kinematic equations:

1. First Equation of Motion \[ v = u + at \]

2. Second Equation of Motion: \[ s = ut + \frac{1}{2} a t^2 \]

3. Third Equation of Motion: \[ v^2 = u^2 + 2as \]

These equations allow us to relate position, velocity, and acceleration to time, making them invaluable tools for problem-solving.

Graphical Interpretation

In uniformly accelerated motion, different graphs can portray the relationship between parameters:

  • Velocity-Time Graph: A straight line indicates uniform acceleration, and its slope is equal to the value of the constant acceleration.
  • Displacement-Time Graph: This graph is a parabola, highlighting the quadratic nature of the relationship between displacement and time.

Examples of Uniform Acceleration

Understanding uniform acceleration conceptually often benefits from specific examples. Below are points illustrating scenarios where uniform acceleration is observed or can be a reasonable approximation.

Falling Objects in a Vacuum

  • Context: An object falling freely near the Earth’s surface in the absence of air resistance.
  • Acceleration: The gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \) is constant.
  • Example Calculation: If an object starts from rest, its velocity after 2 seconds would be \( v = gt = 9.81 \times 2 = 19.62 \, \text{m/s} \).

Cars Accelerating from Rest

  • Context: A car accelerates from rest to reach a certain speed on a straight road.
  • Acceleration: Constant force from the engine results in constant acceleration.
  • Example Calculation: With \( a = 2 \, \text{m/s}^2 \) and starting from rest (\( u = 0 \)), the car’s velocity after 3 seconds would be \( v = at = 2 \times 3 = 6 \, \text{m/s} \).

Inclined Plane without Friction

  • Context: An object sliding down an inclined plane without friction.
  • Acceleration: The component of the gravitational force along the incline provides a constant acceleration.
  • Example Calculation: If the incline is at \( 30^\circ \) and neglecting friction, the acceleration \( a = g \sin(30^\circ) = 4.905 \, \text{m/s}^2 \).

Spacecraft Thrust

  • Context: A spacecraft in outer space applying constant thrust.
  • Acceleration: Due to the constant thrust and lack of air resistance in space, the acceleration remains constant.
  • Example Calculation: With \( a = 0.5 \, \text{m/s}^2 \), the spacecraft’s velocity after 10 seconds from rest would be \( v = at = 0.5 \times 10 = 5 \, \text{m/s} \).

Elevator Moving Upward

Context: An elevator starts from rest and moves upward with constant acceleration. Acceleration: The motor provides a constant force, and thus, the acceleration is constant. Example Calculation: If \( a = 1 \, \text{m/s}^2 \), the elevator’s velocity after 4 seconds from rest would be \( v = at = 1 \times 4 = 4 \, \text{m/s} \).

Each of these examples can be analyzed using the kinematic equations for uniformly accelerated motion. They serve as practical applications of the concept, aiding in your understanding and problem-solving skills.

Questions and Answers

Is gravitational acceleration a form of uniform acceleration.

Yes, in a vacuum, gravitational acceleration is approximately constant near the Earth’s surface.

How can time \( t \) be calculated using the first equation of motion

Time can be determined by rearranging the first equation: \( t = \frac{v-u}{a} \).

Is uniform acceleration applicable in circular motion?

No, because the direction of acceleration changes continuously in circular motion.

By understanding its definitions, characteristics, and associated equations, you lay the groundwork for solving a wide range of problems in physics. Whether you’re observing a car accelerating on a straight road or an object free-falling in a vacuum, the concept of uniform acceleration frequently applies, making it essential for any student or practitioner of physics.

Further Reading

1. Kinematic Equations 2. Uniform Motion  3.  Concept of Acceleration

Related Posts:

Derivative of Velocity

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Related Articles

  • CBSE Class 11 Physics Notes

Chapter 1: Physical World

  • What is Physics? Definition, History, Importance, Scope
  • How is Physics related to Other Sciences?
  • Fundamental Forces

Chapter 2: Units and Measurement

  • System of Units
  • Length Measurement
  • Measurement of Area, Volume and Density
  • Rounding Numbers
  • Dimensional Analysis
  • Significant Figures
  • Errors in Measurement

Chapter 3: Motion in a Straight Line

  • What is Motion?
  • Distance and Displacement
  • Speed and Velocity
  • Acceleration

Uniform Acceleration

  • Sample Problems on Equation of Motion
  • Solving Problems Based on Free Fall
  • Relative Motion
  • Relative Motion in One Dimension
  • Relative Motion in Two Dimension
  • Calculating Stopping Distance and Reaction Time

Chapter 4: Motion in a Plane

  • Scalar and Vector
  • Vector Operations
  • Product of Vectors
  • Scalar Product of Vectors
  • Dot and Cross Products on Vectors
  • Position and Displacement Vectors
  • Average Velocity
  • Motion in Two Dimension
  • Projectile Motion
  • Uniform Circular Motion
  • Centripetal Acceleration
  • Motion in Three Dimensions

Chapter 5: Laws of Motion

  • Contact and Non Contact Forces
  • Inertia Meaning
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Chapter 6: Work, Energy and Power

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Chapter 7: Systems of Particles and Rotational Motion

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  • Introduction to Waves - Definition, Types, Properties
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Uniformly Accelerated Motion or Uniform Acceleration in Physics is a motion in which the object is accelerated at constant acceleration. We have to keep in mind that uniform accelerated motion does not mean uniform velocity i.e. in uniform accelerated the velocity of the object increases linearly with time. An object drop from the top of the roof, a ball rolling downhill, a man jumping from an airplane, etc. are all instances of uniformly accelerated motion. It is crucial to note that due to the interference of gravity or friction, many instances of uniform application do not maintain absolute uniformity of acceleration .

In this article, we will learn about, uniform acceleration, uniform accelerated motion, uniform accelerated motion equations, and others in detail.

Table of Content

What is Uniform Acceleration?

Kinematic equations for uniformly accelerated motion, uniformly accelerated motion in a plane.

  • Practice Questions
If the rate of change of velocity remains constant then the object is said to be in state of Uniform Acceleration. 

As we know acceleration is a vector quantity thus, the direction of motion remains the same in the case of constant acceleration.

Uniform Acceleration Examples

Some Examples of Uniformly accelerated motion are mentioned below:

  • Free-falling object under the force of gravity.
  • A ball rolling down a frictionless slope.
  • A bicycle rolling downhill when the brakes are applied.
  • Angular Acceleration

Uniform Accelerated Motion Formula

There are three formula that are explained by uniform accelerated motion. These formulas are used to explain the motion of an object. The three uniform accelerated motion formula are,

  • s = ut + 1/2at 2
  • v 2 = u 2 + 2as

Uniform Acceleration Graphs

The uniform acceleration graphs is a straight line graphs. The uniform acceleration graphs is added below,

Uniform-acceleration-graph

Non-Uniform Acceleration Graphs

The non-uniform acceleration graph is the graph added below,

Non-uniform-acceleration-graphs

Kinematic Equations for Uniformly Accelerated Motion provide a relation between initial velocity, acceleration, final velocity, distance covered, and time taken by the object. Let’s take the initial velocity of an object as “u”, the constant acceleration applied to the object is “a”, and the final velocity achieved by the object under this acceleration is “v” velocity the time taken is “t” and the distance covered is “s”.

First Equation of Motion (Velocity Equation)

First Equation of Motion or Velocity Equation provides the relation between final and initial velocities, acceleration, and time, it is given by,

v = u + at where, v is the final velocity u is the initial velocity a is the acceleration t is the time taken by object

Second Equation of Motion (Distance Equation)

Second Equation of Motion or Distance Equation provides the relation between initial velocities, distance, acceleration, and time, it is given by,

s = ut + 1/2 at 2 where, s is the distance u is the initial velocity a is the acceleration t is the time taken by the object

Third Equation of Motion

Third Equation of Motion provides the relation between final and initial velocities, distance, and acceleration, it is given by,

v 2 = u 2 + 2as where, s is the distance v is the final velocity u is the initial velocity a is the acceleration

While using these equations sign convention must be followed. If one direction is taken to be positive then the other is taken as negative. A body in free fall can be considered to be in Uniform Accelerated Motion the body in free fall experiences the uniform acceleration of earth’s gravity. Generally, the upward motion of the object is considered to be positive whereas, the downward motion is considered to be negative.

Learn more about, Equation of Motion

Projectile motion is an example of Uniformly Accelerated Motion in a Plane. Projectile motion accelerates uniformly under the force of gravitation. As we know that gravity works only in a vertical direction i.e. in the y-direction the motion in the x-direction is of constant velocity. Thus, we see that projectile motion can be broken into two different motions, and the equation of motion is calculated separately in these two motions.

Uniformly Accelerated Motion is the motion of an object when the acceleration of the object remains constant. It can be in one dimension, two dimensions, or three dimensions.

V-T Graph for Uniform Accelerated Motion

In uniform accelerated motion the velocity of the object changes at constant speed. The graph for the same is added below,

V-T-Graph-for-Uniform-Accelerated-Motion

  • Uniform and Non-Uniform Motion

Examples on Uniform Accelerated Motion

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

Solution: 

Given, u = 15m/s a = 2 m/s 2 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     = 15 + (2)(5)     = 15 + 10     = 25 m/s

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

Given, u = 40m/s a = -5 m/s 2 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     = 30 – (5)(5)     = 30 – 25     = 5 m/s 

Example 3: If a body is moving at an acceleration of -5 m/s 2 . If the initial speed was 40m/s, what will be the speed in 5 seconds? 

Given, u = 40m/s a = -5 m/s 2 t = 5  For finding out the value of “s”, the first equation of motion can be used.  s = ut + 1/2a(t) 2 Plugging the values in this equation, s = ut + 1/2a(t) 2 s = (40)(5) + 1/2(-5)(5) 2 s = 200 + (-125/2) s = 200 – 125/2 s = (400 – 125)/2 = 275/2

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

Given, u = 20 m/s a = 10 m/s 2 t = 2  For finding out the value of “s”, the first equation of motion can be used.  s = ut + 1/2a(t) 2 Plugging the values in this equation, s = ut + 1/2a(t) 2 s = (20)(2) + 1/2(10)(2) 2 s = 40 + 20 = 60 m

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

Solution:   

Given, u = 0 m/s v = 20 m/s 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/s 2 For finding out the distance, a third equation of motion will be used v 2 = u 2 + 2as 20 2 = 0 + 2(10)s 400 = 20s  s = 20 m

Practice Questions on Uniform Acceleration

Q1: If the velocity of a car changes from 60 m/s to 70 m/s in 30 min and 70 m/s to 80 m/s in next 30 min. Verify if the car is in the Uniform Acceleration.

Q2: If a body starts moving at 10 m/s and its uniformly accelerates at 2m/s 2 . Find the distance covered by the body in 30 minutes

Q3: If the initial velocity of a bike is 45 m/s and its final velocity is 80 m/s accelerating uniformly for 10 minutes. Find the acceleration.

FAQs on Uniformly Accelerated Motion

1. define uniform acceleration.

Uniform acceleration is the acceleration that does not vary with time. In such cases, the rate of change of velocity remains constant. 

2. What is Non-Uniform Motion?

An object moving with variable velocity is said to be non-uniform motion. In non uniform motion an object covers unequal distance in unequal time.

3. What is Uniformly Accelerated Motion?

Uniformly accelerated motion is defined by a motion in a straight line with a constant acceleration and zero difference.

4. What are Examples of Uniformly Accelerated Motion?

A ball rolling down a slope, a skydiver jumping out of a plane, a ball dropped from the top of a ladder, and a bicycle with its brakes engaged are all instances of uniformly accelerated motion.

5. Is Uniformly Accelerated Motion Uniform Motion?

No, uniform accelerated motion is not considered to uniform motion as in uniform motion velocity is constant whereas in uniform accelerated motion velocity is not constant.

6. What is Uniform Motion?

The motion of an object in which the object travels equal distance in equal interval of time is called the uniform motion. In uniform motion the velocity of object remains the constant.

7. When will you say a body is in Uniform Acceleration?

A body is said to be in uniform acceleration if its velocity changes at constant rate.

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2.2: Accelerated Linear Motion and Generalization

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If you have understood the idea of taking the area under the velocity vs time graph, then this section would be quite simple to understand. And if you are familiar with basic calculus, it is easy to obtain a general solution for a velocity as the n th degree function of time. However, for the sake of completeness and motivation, it will be covered here.

Accelerated Linear Motion

Accelerated linear motion means that the velocity itself is changing at a constant rate. This rate of change of velocity is called acceleration (denoted by \(\overrightarrow{a}\) ). We obtain the following velocity vs time graph for such motion,

The slope of the graph would be the acceleration (due to definition). To find the area under this curve, we employ 2 methods:

Method 1: Dividing into triangle and rectangle

From the graph, it is obvious that the area under this curve can be divided into ∆BCD and rectangle AOCD.

So, area of AOCD is OA×OC, i.e. \(\overrightarrow{u}\)×t      ∵ \(\overrightarrow{u}\) is initial velocity

Area of ∆BCD is,

1 / 2 ×BD×CD

∴Area=  1 / 2 ×t×(\(\overrightarrow{a}\)t)    ∵ Final velocity will be \(\overrightarrow{a}\)×t

Adding both the areas the final formula is,

\(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a}t^2\)

Remember this result.

Method 2: Integration

We know by the equation of a straight line that

\( \overrightarrow{v} =  \overrightarrow{u} +  \overrightarrow{a}t \)    ∵ y=mx+c

From it, we obtain the value of the velocity at that instant of time. So, by integrating v with respect to time,

\(\int v \cdot dt = \int (\overrightarrow{u} + \overrightarrow{a}t) \cdot dt \)

We obtain the general solution,

\(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a}t^2 + s_0\)      ∵ s 0 is the constant of integration. For this equation, it represents the initial displacement

Generalizing the result

Using the equation for velocity frees us up from the need to draw a graph. Because of this, we can create a general equation that can give us the displacement for n th degree time-dependent velocity. Let's take the example of increasing acceleration. The rate of change of acceleration is called jerk ( j ). So,

\( \overrightarrow{a} =  \overrightarrow{a_0} +  \overrightarrow{a}t \)    ....(i)

\( \overrightarrow{v} =  \overrightarrow{u} +  \overrightarrow{a}t \)    ....(ii)

By using (i) in (ii),

\(\overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a}_0 t + \int \overrightarrow{j} \cdot dt \)    ....(iii)

Plugging this into our previous formula for displacement and integrating we get,

\(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a_0}t^2 + \tfrac{1}{6} \overrightarrow{j}t^3 \)

From this, we can see an obvious pattern. Mathematically, it is expressed as,

\( \Sigma_{i=0}^{n} \frac{x_i}{i!}t^i \nonumber\)    

Where 'x i ' represents the initial value of the i th derivative of displacement. This formula can be used to solve any kinematics equation, but the math can turn quite hairy.

Or, if you are familiar with Taylor series, we can simply plug displacement into the equation to obtain:

\( \Sigma_{i=0}^{n} \frac{x_i}{i!}t^i \nonumber\)

This shows the power of knowing a little bit of calculus. Instead of integrating based on a graph, we were directly able to obtain the same definition of displacement. 

Ending note

These derivations were all based on the simple formula, \(\overrightarrow{v} = \dfrac{\Delta \overrightarrow{s}}{\Delta \overrightarrow{t}}\)

With the help of some basic calculus, we were able to obtain the general equation for displacement. But the dependence of these equations on time can sometimes make the problem-solving process lengthy. So, we shall now derive a time-independent equation for uniformly accelerated linear motion.

\(\dfrac{\overrightarrow{v}-\overrightarrow{u}}{\overrightarrow{a}} = t \)    ....(i)

\(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a}t^2\)    ....(ii)

Using (i) in (ii) and simplifying, we obtain,

\(v^2 - u^2 = 2as \)

Notice that we haven't marked any of the values in the final formula as vectors. This is because for vectors, multiplication in the regular sense is meaningless (what does it mean to multiply two directions?). Therefore, we only use the magnitude of the vectors. More precisely, we use the dot product of the vectors. This shall be further discussed under vectors.

Though we have derived these formulae for only linear motion, because of vectors, they can be used in any scenario analyzing uniform acceleration .

Once again, the equations are:

i) \(\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t \)

ii) \(\overrightarrow{s} = \overrightarrow{u}t + \tfrac{1}{2} \overrightarrow{a}t^2 + s_0\)

iii) \(v^2 - u^2 = 2as \)

The generalization of equation (i) and (iii) is similar to the generalization of (ii). With knowledge of integration, it is trivial and is left as an exercise for the reader.

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  3. Uniformly accelerated motion

    problem solving uniformly accelerated motion

  4. The Derivative and Uniformly Accelerated Motion Equations

    problem solving uniformly accelerated motion

  5. Uniformly Accelerated Motion Examples

    problem solving uniformly accelerated motion

  6. uniformly accelerated motion problem solving

    problem solving uniformly accelerated motion

VIDEO

  1. Ch 2 part 2 (motion at constant acceleration )

  2. Physics. Problem solving. 01_12

  3. Lecture 4 || Uniformly Accelerated Motion, Equations of Motion || Foundation || Physics

  4. Uniformly Accelerated Motion: Problem Solving (horizontal axis)

  5. | Uniformly accelerated motion|| @ethioSIMPLEedu

  6. Chapter 7 Motion Lecture 29 Uniformly accelerated motion (Class 9th Physics / Science)

COMMENTS

  1. Uniform Acceleration Motion: Problems with Solutions

    Solution to Problem 1 Problem 2: With an initial velocity of 20 km/h, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 2 A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds.

  2. Uniform acceleration motion Problems and Solutions

    Answer: Given: time t = 4,0 s, initial velocity vi = 21,0 m/s, and finale velocity vf = 35,0 m/s, Find: acceleration a,? and distance car, d? we use formulas to determine the magnitude of acceleration vf = vi + at 35 m/s = 21 m/s + a (4 s) a = 3,5 m/s2 we use formulas to determine the magnitude of distance vf2 = vi2 + 2 ad

  3. 6.2: Uniformly Accelerated Motion

    Classical Mechanics (Tatum) 6: Motion in a Resisting Medium

  4. Kinematic Equations: Sample Problems and Solutions

    Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall? See Answer See solution below. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled. See Answer

  5. Uniform Acceleration

    Initial velocity ( u u) and initial position ( s0 s 0) are essential parameters for problem-solving in uniformly accelerated motion. These conditions act as the starting point for any calculations involving kinematic equations. Idealized Conditions

  6. PDF Solutions for Uniformly Accelerated Motion Problems WorksheetsCh6

    Solutions for Uniformly Accelerated Motion Problems Worksheets Worksheet: POSITION, VELOCITY, AND ACCELERATION (6.3) For each problem, you must draw graphs and give equations for a(t), v(t) and s(t). 1. a) a(t) = 1.2, v(t) = 1.2t, s(t) = 0.6 t2 Draw graphs indicating scale, important points and proper shape.

  7. Uniformly-Accelerated Motion

    Kinematic equations or uniformly accelerated equations are used to solve problems involving constant acceleration. The uniformly accelerated motion equations are the following: (1) v =...

  8. Uniformly Accelerated Motion

    Uniformly Accelerated Motion Directions: On this worksheet you will practice solving problems by using the five basic kinematics equations for uniformly accelerated motion.

  9. PDF 1.6 Solving Uniform Acceleration Problems

    1.6 Solving Uniform Acceleration Problems This highly mathematical section presents the derivations and applications of the equations involving all of the variables related to uniformly accelerated motion. An activity in which students determine their reaction time provides a break from the algebraic problem solving. Expectations Addressed

  10. PDF 1.6 Solving Uniform Acceleration Problems

    Physics Chapter 1 1.6 Solving Uniform Acceleration Problems Now that you have learned the definitions and basic equations associated with uniform acceleration, it is possible to extend your knowledge so that you can solve more complex problems.

  11. PDF 0016 Lecture Notes

    Uniformly Accelerated Motion (UAM) is motion of an object where the acceleration is constant. In other words, the acceleration remains uniform; the acceleration is equal to a number and that number does not change as a function of time. A ball rolling down an incline. A person falling from a plane. A bicycle on which you have applied the brakes.

  12. PDF Uniformly Accelerated Motion Super Problem

    Solution: 1. viy = +25 m/s 2. ay = −9.8 m/s2 3. The final velocity is v v fy a iy y v v fy iy 4. The final velocity is t a t 25 m/s ( 9 .8m/s 2 )(2s) 5 .4 m/s y v v a

  13. Uniformly Accelerated Motion (2/2): Problem-Solving Example

    NOTE: The second-to-last equation from the bottom should read vf^2 = vi^2 + 2aΔx, not vf = ... In that equation, both of the velocities are squared. The vide...

  14. Using equations of motion (1 step numerical)

    AboutTranscript. In this video, we will solve 2 numerical on uniformly accelerated motion by using the three equations of motion (kinematic equations) v = u+at, s = ut + 1/2 at^2 and v^2 = u^2+2asWe will calculate the time taken in the first numerical and distance in another. Created by Mahesh Shenoy.

  15. Solving Problems Using Kinematic Equations (Uniformly Accelerated Motion)

    Learn how to solve motion problems in Physics using kinematic equations used in uniformly acceleration motion, i.e. acceleration is assumed constant.Watch ho...

  16. Uniformly Accelerated Motion (Part I ) Horizontal Motion ...

    Detailed explanation of uniformly accelerated motion and the use of the 4 kinematic equations.

  17. Uniformly Accelerated Motion: Exploring the Dynamics of Speed and

    ad the branch of physics that deals with the description of motion without considering the forces causing the motion. By understanding uniformly accelerated motion, we can analyze and predict the behavior of objects in motion using the equations of motion.

  18. Uniformly Accelerated Motion Calculator

    The Uniformly Accelerated Motion calculator or (kinematic equations calculator) solves motion calculations involving constant acceleration in one dimension, a straight line. It can solve for the initial velocity u, final velocity v, displacement s, acceleration a, and time t.

  19. How to solve uniform motion problems

    The distance formula for uniform motion problems. In this lesson we'll look at how to compare and solve for different values in the ???\text{Distance}=\text{Rate} \cdot \text{Time}??? equation when you have related scenarios. Uniform motion explains the distance of an object when it travels at a constant speed, the rate, over a period of time.

  20. Physics 71 Day 5 [Part 2/8]

    In this video, we gave a guide and some strategies on how to solve problems involving kinematics of uniformly accelerated motion.

  21. Uniform Accelerated Motion: Equations, Graphs, and Examples

    Uniformly Accelerated Motion or Uniform Acceleration in Physics is a motion in which the object is accelerated at constant acceleration. We have to keep in mind that uniform accelerated motion does not mean uniform velocity i.e. in uniform accelerated the velocity of the object increases linearly with time.

  22. 2.2: Accelerated Linear Motion and Generalization

    With the help of some basic calculus, we were able to obtain the general equation for displacement. But the dependence of these equations on time can sometimes make the problem-solving process lengthy. So, we shall now derive a time-independent equation for uniformly accelerated linear motion.