These equations give a relationship between initial velocity, final velocity, time taken, acceleration and distance travelled by the bodies.
A body having an initial velocity 'u' acted upon by a uniform acceleration 'a' for time 't' such that final velocity of the body is 'v'.
v – u = at v = u + at
Where ; v = final velocity of the body u = initial velocity of the body
a = Acceleration t = time taken
It gives the distance travelled by a body in time t.
A body having an initial velocity 'u' acted upon by a uniform acceleration 'a' for time 't' such that final velocity of the body is 'v' and the distance covered is 's'.
The average velocity is given by :
distance covered = average velocity × time
but v = u + at (from first equation of motion)
Thus,
Where ; s = distance travelled by the body in time t
u = initial velocity of the body
a = Acceleration
t = time taken
A body having an initial velocity 'u' moving with a uniform acceleration 'a' for time 't' such that final velocity 'v' and the distance covered is 's'. the third equation of motion is v^{2} = u^{2 }+ 2at. it gives the velocity acquired by a body in travelling a distance s.
v = u + at .....(i)
Squaring eq. (i), we have
v^{2} = (u + at)^{2}
Substituting the value of eq (ii) in eq.(iii), we get. v^{2} = u^{2} + 2as
► View AnswerImportant Note:
Velocity in m/s = _____ × velocity in km/h.
36 km h^{1} = 36000 m h^{1}
To convert km h^{1 } to ms^{1}, multiply by 5/18.
To convert ms^{1}^{ }to km h^{1}^{ }, multiply by 18/5.
1 km = 1000 m
1m = 100 cm = 1000 mm
Important Points for solving Numericals
 Distance in kilometres should be converted into metre.
 Before solving problems, ensure that the data provided have the same system of unit, i.e. either they should be in SI system or CGS system.
 If a body start from rest, its initial velocity (u) is zero, (u =0)
 If a moving body comes to rest/stops, its final velocity (v) is zero, (v =0)
 If a body is moving with uniform velocity, its acceleration is zero, (a =0)
 If the body is falling under gravity, the value of acceleration should be taken g.
A graph is a line, straight line or curved, showing the relation between two variable quantities of which one varies as a result of the change in the other.
The quantity which changes independently is called independent variable and the one which changes as a result of the change in the other is called dependent variable.
Plotting a Graph
Uses of Graphs
Competitive Window

This graph is plotted between the time taken and the distance covered, the time is taken along the xaxis and the distance covered is taken along the yaxis.
Speed = distance/time
When the body is at Rest
When position of the body does not change with time then it is said to be stationary, the distancetime graph of such a body is a straight line parallel to xaxis.
Distancetime graph for a Stationary Body
When the body is in Uniform Speed
When the position of the body changes by equal intervals of time then body is said to be moving with uniform speed. The distancetime graph of such a body is a straight line, inclined to xaxis.
Special CaseI
In uniform motion along a straight line the position x of the body at any time t is related to the constant velocity as,
x_{A} = vt Starting form zero
x_{B} = x_{0} + vt starting from x_{0}
Special CaseII
Slope of line A = tanθ_{A} = tan0 (∵ θ_{A} = 0)
=> zero velocity
Slope of line B = tanθ_{B} = positive velocity
Slope of line C = tanθ_{c} = more positive velocity
∵ θ_{C}> θ_{B} (tanθ_{C}> θ_{B })
Then v_{C}> v_{B}
Slope of line D = tan (θ_{D}) = negative velocity.
When the body is in Motion with a Nonuniform (or Variable) Speed
DistanceTime graph for a body moving with nonunifrom speed.
The positiontime graph is not a straight line, but is a curve.
The speed of the body at any point is known as instantaneous speed and can be calculated by finding the slope at that point.
So instantaneous speed of the body at point A.
Slope at point A = tanθ_{A} = AE \ CE
instantaneous speed of the body at point B
Slope at point B = tanθ_{B} = BF\DF
θ_{B}> θ_{A} so slope at point B is greater than the slope at point A.
Hence speed of body at point B is a greater than, the speed of body at point A.
When the Speed Decreases with Passage of Time
Slope at point A > slope at point B (θ_{A}> θ_{B})
So, speed at point A > speed at point B
Important Note: A distance time graph can never be parallel to yaxis (representing distance) because this line has slope of 90° and slope = tanθ = tan90° = infinite, which means infinite speed. It is impossible.
Acceleration from Displacementtime Graph
For line A : A straight line displacementtime graph represents a
uniform velocity and zero acceleration
for line B : A curved displacementtime graph rising upward represents an increasing velocity and positive acceleration
For line C : A curved displacementtime graph falling downwards, represents a decreasing velocity and negative acceleration.
The variation in velocity with time for an object moving in a straight line can be represented by a velocitytime graph. In this graph, time is represented along the xaxis and velocity is represented along the yaxis.
Acceleration = (speed or velocity over time)
hence the slope of the speed/velocitytime graph, gives the acceleration of the body.
Distance = speed × time
hence, area enclosed between the speedtime graph line and xaxis (time axis) gives the distance covered by the body. Similarly area enclosed between the velocitytime graph line and the xaxis (time axis) gives the displacement of the body.
Note: Since the graph takes into account, only the magnitude hence velocitytime graph is not different from speedtime graph.
When the body is moving with Constant Velocity
When the body moves with constant velocity i.e. its motion is uniform.
The speed or velocity of the body is uniform hence the magnitude remains same. The graph is a straight line parallel to xaxis (timeaxis). Since the velocity is uniform. Its acceleration is zero. The slope of the graph in this case is zero.
Conclusion: Velocitytime graph of a body moving with constant velocity is a straight line parallel to time axis.
When the body is moving with Uniform Acceleration
The speed or velocity is changing by equal amounts in equal interval of time, the speed or velocity time graph of such a body is a straight line inclined to xaxis (timeaxis).
When the body is moving with a Nonuniform (or Variable) Acceleration
The speed or velocitytime graph is not a straight line but is a curve.
The line has different slopes at different times, its acceleration is variable. At point A, slope is less hence acceleratoin is less. At point B slope is more hence acceleration is more.
Note: Speed or velocitytime graph line can never be paralled to yaxis (speed axis), because slope angle becomes 90° than tan90° is infinite it is impossible.
As distance or displacement = speed or velocity x time, hence the distance or displacement can be calculated from speed or velocitytime graph.
When Speed or Velocity is Uniform (or Constant)
Distance/displacement = Area of rectangle ABCD = AB × AD
Thus, We find that the area enclosed by velocitytime graph and the time axis gives the distance travelled by the body.
When Acceleration is Uniform (or Constant)
distance or displacement = area of right triangle OAB =
When Speed or Velocity as well as Acceleration is Nonuniform (or Variable).
The speedtime graph of a body moving irregularly with variable speed and acceleration. For a small interval of time Δt, as there is not much change in speed, hence the speed can be taken as constant.
For this small time interval.
Distance Δs = vΔt = Area of the blackened strip.
For whole time interval between t_{1} and t_{2}, distance = sum of areas of all the strips, between t_{1} and t_{2} = area of shaded figure ABCD.
A number of useful results can be deduced from velocity time graph.
Competitive Window
Represents a velocitytime graph BC, in which AB represents the initial velocity u, CE represents final velocity v, such that the change in velocity is represented by CD, which takes place in time t, represented by AE.
Acceleration = slope of the graph line BC.
v – u = at
v = u + at
Distance travelled = Area of trapezium ABCE
= Area of rectangle ABDE + Area of triangle BCD
From the velocitytime graph distance covered = Area of trapezium ABCE
Substituting the value of t in eq (i)
∵ A^{2} – B^{2} = (A + B) × (A – B)
16 videos69 docs56 tests

1. What are the equations of uniformly accelerated motion? 
2. What is the graphical representation of motion? 
3. How can we use graphical method to find equations of motion? 
4. What is the significance of equations of motion in physics? 
5. What are some reallife applications of uniformly accelerated motion? 
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