Motion in a Straight Line

Kinematic Equations for Uniformly Accelerated Motion

Suppose you are driving a vehicle on a road. The velocity of the vehicle is never uniform. Its velocity increases and decreases randomly. As you move the velocity of the vehicle on the road keeps on changing and this is said to be accelerated motion. Let us study the different kinematic equations of accelerated motion.

Suggested videos

Play
Play
Play
previous arrow
next arrow
previous arrownext arrow
Slider

A body travels in a uniform motion if it moves with a constant velocity, i.e it undergoes equal displacements in equal intervals of time. If the magnitude or direction of the velocity changes, then the body is not in uniform motion. If a body is in uniform motion, then the net external force on the body is zero. The inverse of this statement is also true (if the net external force on the body is zero, then the body is in uniform motion).

Uniform Acceleration

A body under constant acceleration is uniformly accelerated motion. This is when a constant external force is applied to the body. The direction of acceleration is very important in the change of velocity. However, the direction of the acceleration of a body may not be the same as the direction of its motion. For example, if a car travels at a speed of 60 km/hour then it will cover a distance of 1 km/minute. Here the motion of car acceleration is uniform.

Browse more Topics under Motion In A Straight Line

What is Acceleration?

Kinematic Equations of Motion

If an object starts with velocity ”u” and after some time “t” its velocity changes to v, if the uniform acceleration is a and distance traveled in time (t) is s, then we obtain the following kinematic equations of uniformly accelerated motion.

First Equation Of Motion

Let an object is moving with uniform acceleration
u = initial velocity of the object
v = final velocity of object
a = uniform acceleration
Let object reach point B after time (t) Now, from the graph

Slope= Acceleration(a)=  \( \frac{change in velocity}{time} \)

Change in velocity = AB= \(\vec{v}-\vec{u}\)

Time = AD = t

a =  \( \frac{\vec{v}-\vec{u}}{t} \)

Solving this we get the first equation of motion:

\( \vec{v} = \vec{u} + \vec{a}t \)

Learn more about Relative Velocity Motion in Two Dimensions here.

Second Equation Of Motion

Consider an object starting with an initial velocity u and moving with uniform acceleration a. Distance covered by the object at the given time t is given by the area of the trapezium ABDOE. Let in the given time ‘t’, the displacement covered by the moving object  ‘s’ is given by the area of a trapezium, ABDOE

Displacement (s) = Area of ABD + Area of ADOE

\( \frac{1}{2} \) × AB× AD+ AE ×OE, AB= dc= at

= \( \frac{1}{2} \) at× t+ ut

Hence, the second equation of motion is:

\( \vec{s}= \vec{u}t+ \frac{1}{2} \vec{a}t²\)

Velocity time graph of a uniformly accelerated motion

The velocity-time graph of a uniformly accelerated motion is a straight line graph inclining towards the time axis. If the object has positive constant acceleration, the graph slopes upward. In case the object has negative constant acceleration, the velocity-time graph will slope downward.

Learn more about Instantaneous Velocity and Speed here.

Solved Examples For You

Q. An automobile traveling with a speed of 60 Km/h can apply the brake to stop within a distance of 20 m If the car is going twice as fast i.e., 120 Km/h the stopping distance will be

  1. 60m
  2. 40m
  3. 20m
  4. 80m

Answer: D

Q2. A cart that is free to move in one dimension is moving in the positive direction and slowing down under the influence of a constant backward pull until the cart comes to a momentary stop and reverses direction. What are the signs of the cart’s velocity and acceleration during the moment it is stopped?

  1. velocity : 0 acceleration : +
  2. acceleration : 0 velocity : +
  3. velocity : 0 acceleration : –

Answer: C. Since the cart has stopped at the moment, the speed and velocity of the cart are zero. Hence the cart achieves negative velocity which means that it should be accelerating in the negative direction to raise velocity from zero to finite value in the negative direction.

Share with friends

Customize your course in 30 seconds

Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
tutor
tutor
Ashhar Firdausi
IIT Roorkee
Biology
tutor
tutor
Dr. Nazma Shaik
VTU
Chemistry
tutor
tutor
Gaurav Tiwari
APJAKTU
Physics
Get Started

Leave a Reply

Your email address will not be published. Required fields are marked *

Download the App

Watch lectures, practise questions and take tests on the go.

Customize your course in 30 seconds

No thanks.