# 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.

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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 theÂ velocity changes, then the body is not in uniform motion. If a body is in uniform motion, then the net externalÂ force on the body is zero. The inverse of this statement is also true (ifÂ the net externalÂ force on the body is zero, then theÂ 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 asÂ 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 isÂ 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)=Â  $$\frac{change in velocity}{time}$$

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

a =Â  $$\frac{\vec{v}-\vec{u}}{t}$$

Solving this we get the first equation of motion:

$$\vec{v} = \vec{u} + \vec{a}t$$

### 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 objectÂ  ‘s’ is given by the area of a trapezium, ABDOE

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

$$\frac{1}{2}$$ Ã— ABÃ— AD+ AE Ã—OE, AB= dc= at

=Â $$\frac{1}{2}$$ atÃ— t+ ut

Hence, the second equation of motion is:

$$\vec{s}= \vec{u}t+ \frac{1}{2} \vec{a}tÂ²$$

#### 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.

## Solved Examples For You

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

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

Q2.Â 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 accelerationÂ : +
2. accelerationÂ : 0Â velocity : +
3. velocity : 0 accelerationÂ : –

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.

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