Videos Related with Equation of Motion of Uniform Acceleration

Note on Equation of Motion of Uniform Acceleration

  • Note
  • Things to remember
  • Videos
  • Exercise
  • Quiz

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.

  • 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
.

Very Short Questions

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.

0%
  • In SI system, the unit of speed is ______.

    m/s


    cm/s


    km/hr


    km/min


  • Velocity of car cannot be changed by ______

    Keeping speed and direction constant


    changing speed and direction


    changing the speed


    Changing the direction


  • What does a horizontal line in a distance-time graph indicate?

    body at non-uniform speed
    body at straight motion
    body at flight
    body at rest
  • Which type of distance-time graph is impossible?

    curved line
    straight line
    horizontal line
    Vertical line
  • What does a horizontal line in a speed-time graph indicate?

    Constant speed
    constant acceleration
    Body at non-uniform speed
    body at flight
  • A book lying on a table follows which laws of Newton's Motion

    4th law
    3rd law
    2nd law
    1st law
  • When a ball bounces off a wall, which law of Newton's Motion does it follow?

    4th law
    2nd law
    1st law
    3rd law
  • When a tablecloth of a table with a plate is swiftly pulled and a plate remains intact in its position which law of Newton's Motion does it follow?

    4th law
    1st law
    3rd law
    2nd law
  • You scored /8


    Take test again

DISCUSSIONS ABOUT THIS NOTE

No discussion on this note yet. Be first to comment on this note