Subject: Science

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

=

Or,

Or, at = v-u

∴ v= u + at ........... (i)

This is the first equation of motion.

Or,

Or,

As both equations are equal

Or,

Or, 2s = (u + v) × t

∴........(ii)

This is the second equation of the motion.

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.

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

\(\therefore\) v^{2} = u^{2} + 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\(t^2\)
- v
^{2}= u^{2}+ 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.

A motorbike of mass 500kg moving with a velocity 72km/hr travels a distance of 1.2km with acceleration of 0.7 m/s^{2}. Calculate the final velocity and time taken to cover the distance.

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/s^{2}

By using formula,

v^{2 }= u^{2} + 2as

= 20^{2} + 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/s^{2}

Time(t) = ?

Now, a = or, 3 =

or. t = 24/3 or, t = 8s

It takes 8 seconds to gain the velocity of 24m/s.

Here,

Initial velocity (u) = 0

Final velocity (v) = 24m/s

Acceleration (a) = 3m/s

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/s^{2}

Retardation is the negative of acceleration,

So retardation = 5m/s^{2}. Similarly,

s = ut + ½ at^{2}

or, s = 20 x 4 + ½ (-5) x 42

or, s = 80 – 40 = 40m Hence, the retardation is 5m/s^{2} and the car will travel 40m distance before it comes to rest.

A bus is moving with a velocity of 72km/hr on seeing the red signal of traffic 27m ahead on the road, the driver applied brakes and the bus stopped at a distance of 25m. What is the retardation and how long does the bus take to come in 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,

v^{2} = u^{2 }+ 2as

or, 0 = 20^{2} + 2as

or, -400 = 50a

or, a = -8m/s^{2}

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/s^{2}

Now,

Height(h) = ut + ½ gt^{2}

= 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/s^{2}

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.

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/s

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.

^{}A car starts from rest and accelerates uniformly to 21m/s in a distance of 21m. Calculate (a) The acceleration produced (b) Time taken by the car to travel that distance.

Solution:

Given,

Initial velocity (u) = 0

Final velocity (v) = 21m/s

Distance (s) = 21m

Acceleration (a) = ?

By using formula,

v2 = u^{2} + 2as

or, 21^{2} = 0^{2} + 2a x 21

or, 441 = 42a

or, 441/42 = a

or, a = 10.5m/s^{2} (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/s^{2} 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/s^{2}

Thus, the engine of the micro gains 0.55m/s^{2} acceleration.

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/s

Thus, the engine of the micro gains 0.55m/s

Solution:

Initial velocity (u) = 54km/hr = 54000/3600 m/s = 15m/s

Final velocity (v) = 0

Retardation (a) = 2.5m/s^{2}

Acceleration (a) = -2.5m/s^{2}

Distance travelled (s) = ?

Now,

v^{2} = u^{2} + 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/s^{2} 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.

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/s

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.

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 at^{2}

Using (ii) in (i)

s = x t

= x t

=

= ut + ½ at

∴ s = ut + ½ at^{2},

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 = v^{2 }– u^{2}

v^{2} = u^{2} + 2as

proved.

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