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Equation of Motion of Uniform Acceleration

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Equations involving displacement, initial velocity, final velocity, acceleration and time of motion of a moving body are equations of motion. This note gives us the information about equation of motion of uniform acceleration and its derivation.
Equation of Motion of Uniform Acceleration

Equations involving displacement, initial velocity, final velocity, acceleration and time of motion of a moving body are equations of motion.

Consider a body moving in a straight line with uniform acceleration as shown in the figure.

Let,
Displacement = s
Initial velocity = u
Final velocity = v
Acceleration = a
Time taken = t

Relation between u, v, a, and t


=
Or,
Or, at = v-u
∴ v= u + at ........... (i)
This is the first equation of motion.

Relation between s, u, v and t


Or,

Or,
As both equations are equal
Or,
Or, 2s = (u + v) × t
∴........(ii)
This is the second equation of the motion.

Relation between s, u, a and t

We already have,
V= u + at ........(i)
.......(ii)
Putting value of v from equation (i) in (ii)
Or, s =
Or, s = × t
Or, s = (2u × t + at × t )\(\frac{1}{2}\)
∴ s = ut + \(\frac{1}{2}\)a\(t^2\) ........(iii)
This is the third equation of motion.

Relation between u, v, a and s

We have,
v = u + at..............(i)
s = \(\frac{u +v}{t}\)× t
Putting the value of t from (i) in the equation (ii),
s = \(\frac{v+ u}{2}\) \( \times \frac{v - u}{a}\)

Or, s = \(\frac{v^2- u^2}{2a}\)

\(\therefore\) v2 = u2 + 2as..................(iv)

This is the fourth equation of motion.

Things to remember
  • Equations involving displacement, initial velocity, final velocity, acceleration and time of motion of a moving body are equations of motion.
  • v= u + at is the first equation of motion.
  •  s = ut +  \(\frac{1}{2}\)a\(t^2\)
  • v2 = u2 + 2as
  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Equation of Motion of Uniform Acceleration
Derivation of Equations of Motion
Equation of motion
The Equations of Motion
Questions and Answers

Solution:
Mass (m) = 500kg
Initial velocity (u) = 72km/hr = = 20m/s
Final velocity (v) = ?
Distance travelled (s) = 1.2km = 1200m
Time (t) = ?
Acceleration (a) = 0.7 m/s2
By using formula,
v2 = u2 + 2as
= 202 + 2 x 0.7 x 1200 = 2080
or, v = = 45.60 m/s and ,
t = (v-u)/a = (45.6-20)/0.7 = 36.57 sec
Hence, the final velocity and time are 45.6m/s and 36.57 s respectively.

Solution:
Here,
Initial velocity (u) = 0
Final velocity (v) = 24m/s
Acceleration (a) = 3m/s2
Time(t) = ?
Now, a = or, 3 =

or. t = 24/3 or, t = 8s
It takes 8 seconds to gain the velocity of 24m/s.

Solution:
Initial velocity (u) = 72km/hr = = = 20m/s Final velocity (v) = 0
Time (t) = 4s
Retardation (a) = ?
Distance travelled (s) = ?
As we know that,
a =
or, a = -20/4
or, a = -5m/s2
Retardation is the negative of acceleration,
So retardation = 5m/s2. Similarly,
s = ut + ½ at2
or, s = 20 x 4 + ½ (-5) x 42
or, s = 80 – 40 = 40m Hence, the retardation is 5m/s2 and the car will travel 40m distance before it comes to rest.

Solution:
Here, Initial velocity (u) = 72km/hr = = = 20m/s
Final velocity (v) = 0
Distance travelled(s) = 25m
Retardation (a) = ?
Time (t) = ?
By using formula,
v2 = u2 + 2as
or, 0 = 202 + 2as
or, -400 = 50a
or, a = -8m/s2
Similarly,
t = (v-u)/a = (0-20)/-8 = 2.5 sec

Solution:
Here,
Initial velocity (u) = 0
Time (t) = 3s
Height of the tower(h) = ?
Acceleration due to gravity (g) = 9.8 m/s2
Now,
Height(h) = ut + ½ gt2
= 0 x 3 + ½ x 9.8 x 32
=44.1m
Hence, the height of the tower is 44.1m.

Solution:
Given,
Initial velocity (u) = 20m/s
At maximum height, v = 0
Maximum height (h) = ?
Time taken (t) = ?
Acceleration due to gravity (g) = 9.8 m/s2
Here, when the object is thrown up -g =
Or, -9.8 = -
Or, t = 2.04 sec To return to the original position, the time required = 2 x t = 4.08 sec.

Solution:
Given,
Initial velocity (u) = 0
Final velocity (v) = 21m/s
Distance (s) = 21m
Acceleration (a) = ?
By using formula,
v2 = u2 + 2as
or, 212 = 02 + 2a x 21
or, 441 = 42a
or, 441/42 = a
or, a = 10.5m/s2 (b) Time taken (t) = ?
Since, a = or, 10.5 = 21/t
or, t = 21/10.5
or, t = 2 sec
Hence, acceleration of the car is 10.5m/s2 and time taken to travel the distance is 2 sec.

Solution:
Given,
Initial velocity (u) = 0
Final velocity (v) = 60km/hr = 60000/3600 m/s = 16.67 m/s
Time taken (t) = 30s
Acceleration (a) = ?
By using the relation, a =
or, a =
or, a = 0.55 m/s2
Thus, the engine of the micro gains 0.55m/s2 acceleration.

Solution:
Initial velocity (u) = 54km/hr = 54000/3600 m/s = 15m/s
Final velocity (v) = 0
Retardation (a) = 2.5m/s2
Acceleration (a) = -2.5m/s2
Distance travelled (s) = ?
Now,
v2 = u2 + 2as or, 0 = 152 + 2 x (-2.5) x s
or, 0 = 225 – 5s
or, 5s = 225
or s = 45m
Here, a child is 50m away from the vehicle. So, it does not cross out the child.

Solution:
Here,
Mass(m) = 7kg
Force applied (F) = 24N
Distance travelled (s) = 20m
Time taken (t) = ?
Initial velocity (u) = 0
Here,
F = ma
Or, a =
Or, a =
Or, a = 3.42m/s2 Similarly,
v2 = u2 + 2as
or, v2 = 0 + 2 x 3.42 x 20
or, v = 11.69 m/s Again,
a =
or, 3.42 =
or, t = 11.69/3.42
= 3.41 sec Hence, the body takes 3.41 sec to come at rest.
In the given figure, balanced force is acting on the body.
Balanced force is a number of forces acting on a body which does not change its state of rest or uniform motion.

Suppose while moving from A to B,
Initial velocity = u
Final velocity = v
Time taken = t
Acceleration = a
According to the definition of acceleration,
a =
or, at = v – u
or, u + at = v
Hence, v = u + at, proved.

Suppose while moving from A to B,
Initial velocity = u
Final velocity = v
Time taken = t
Distance travelled = s

As we know that,
Average velocity =
or, s/t =
s = x t

Let us suppose that,
Initial velocity at A = u
Final velocity at B = v
Distance travelled = s
Acceleration produced = a
As we have ,
s = x t ……(i)

and,
v = u + at ……(ii)
Using (ii) in (i)
s = x t
= x t
=
= ut + ½ at2
∴ s = ut + ½ at2,
proved.

When an object is moving from A to B with an initial velocity 'u', and final velocity 'v' , so that the acceleration produced is 'a' and the distance travelled is 's'
We have, S = x t
= x
=
Or, 2as = v2 – u2
v2 = u2 + 2as

proved.
Quiz

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