Please Wait... While Loading Full Video### EQUATION OF MOTION AND UNIFORM CIRCULAR MOTION

`color{red} ♦` EQUATION OF MOTION BY GRAPHICAL METHOD

`color{red} ♦` UNIFORM CIRCULAR MOTION

`color{red} ♦` UNIFORM CIRCULAR MOTION

When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion.

There are three such equations. These are:

`v = u + a t` ........(8.5)

`s = u t + ½ a t^2` ........(8.6)

`2 a s = v^2 – u^2` .........(8.7)

where `u` is the initial velocity of the object which moves with uniform acceleration `a` for time `t, v` is the final velocity, and `s` is the distance travelled by the object in time t. Eq. (8.5) describes the velocity-time relation and Eq. (8.6) represents the position-time relation.

Eq. (8.7), which represents the relation between the position and the velocity, can be obtained from Eqs. (8.5) and (8.6) by eliminating t. These three equations can be derived by graphical method.

`ul"EQUATION FOR VELOCITY-TIME RELATION"`

Consider the velocity-time graph of an object that moves under uniform acceleration as

shown in Fig. 8.8 (similar to Fig. 8.6, but now with `u ≠ 0`). From this graph, you can see that initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t.

The velocity changes at a uniform rate a.

In Fig. 8.8, the perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by `OC. BD = BC – CD`, represents the change in velocity in time interval t.

Let us draw AD parallel to OC. From the graph, we observe that

`BC = BD + DC = BD + OA`

Substituting `BC = v` and `OA = u`,

we get `v = BD + u`

or `B D = v – u` (8.8)

From the velocity-time graph (Fig. 8.8), the acceleration of the object is given by

` a = text ( Change in velocity)/text( time taken)`

` = (BD)/(AD) = (BD)/(OC)`

Substituting `OC = t`, we get

` a = (BD)/t`

or `B D = a t` (8.9)

Using Eqs. (8.8) and (8.9) we get

` v = u + a t`

`ul"EQUATION FOR POSITION-TIME RELATION"`

Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 8.8, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.

Thus, the distance s travelled by the object is given by

`s =` area `OABC` (which is a trapezium)

`=` area of the rectangle `OADC +` area of the triangle `ABD`

`= OA xx OC + 1/2 (AD xx BD)` (8.10)

Substituting `O A = u, O C = A D = t` and `BD = a t`, we get

` s = u xx t + 1/2 (t xx a t)`

or ` s = ut + 1/2 at^2`

`ul"EQUATION FOR POSITION–VELOCITY RELATION"`

From the velocity-time graph shown in Fig. 8.8, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium `OABC` under the graph.

That is, `s =` area of the trapezium `OABC`

` = ( (OA + BC) xx OC)/2`

Substituting `OA = u, BC = v` and `OC = t, ` we get

` s = ( (u + v) t)/2` ........(8.11)

From the velocity-time relation (Eq. 8.6), we get

` t = (v -u )/a` .........(8.12)

Using Eqs. (8.11) and (8.12) we have

` s = ( (v + u) xx (v - u) )/(2a)`

or `2 a s = v^2 – u^2`

There are three such equations. These are:

`v = u + a t` ........(8.5)

`s = u t + ½ a t^2` ........(8.6)

`2 a s = v^2 – u^2` .........(8.7)

where `u` is the initial velocity of the object which moves with uniform acceleration `a` for time `t, v` is the final velocity, and `s` is the distance travelled by the object in time t. Eq. (8.5) describes the velocity-time relation and Eq. (8.6) represents the position-time relation.

Eq. (8.7), which represents the relation between the position and the velocity, can be obtained from Eqs. (8.5) and (8.6) by eliminating t. These three equations can be derived by graphical method.

`ul"EQUATION FOR VELOCITY-TIME RELATION"`

Consider the velocity-time graph of an object that moves under uniform acceleration as

shown in Fig. 8.8 (similar to Fig. 8.6, but now with `u ≠ 0`). From this graph, you can see that initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t.

The velocity changes at a uniform rate a.

In Fig. 8.8, the perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by `OC. BD = BC – CD`, represents the change in velocity in time interval t.

Let us draw AD parallel to OC. From the graph, we observe that

`BC = BD + DC = BD + OA`

Substituting `BC = v` and `OA = u`,

we get `v = BD + u`

or `B D = v – u` (8.8)

From the velocity-time graph (Fig. 8.8), the acceleration of the object is given by

` a = text ( Change in velocity)/text( time taken)`

` = (BD)/(AD) = (BD)/(OC)`

Substituting `OC = t`, we get

` a = (BD)/t`

or `B D = a t` (8.9)

Using Eqs. (8.8) and (8.9) we get

` v = u + a t`

`ul"EQUATION FOR POSITION-TIME RELATION"`

Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 8.8, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.

Thus, the distance s travelled by the object is given by

`s =` area `OABC` (which is a trapezium)

`=` area of the rectangle `OADC +` area of the triangle `ABD`

`= OA xx OC + 1/2 (AD xx BD)` (8.10)

Substituting `O A = u, O C = A D = t` and `BD = a t`, we get

` s = u xx t + 1/2 (t xx a t)`

or ` s = ut + 1/2 at^2`

`ul"EQUATION FOR POSITION–VELOCITY RELATION"`

From the velocity-time graph shown in Fig. 8.8, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium `OABC` under the graph.

That is, `s =` area of the trapezium `OABC`

` = ( (OA + BC) xx OC)/2`

Substituting `OA = u, BC = v` and `OC = t, ` we get

` s = ( (u + v) t)/2` ........(8.11)

From the velocity-time relation (Eq. 8.6), we get

` t = (v -u )/a` .........(8.12)

Using Eqs. (8.11) and (8.12) we have

` s = ( (v + u) xx (v - u) )/(2a)`

or `2 a s = v^2 – u^2`

Q 3264791655

A train starting from rest attains a velocity of `72 km \ \ h^(–1)` in `5` minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.

Class 9 Chapter 8 Example 5

Class 9 Chapter 8 Example 5

(i)We have been given

`u = 0 ; v = 72 km \ \ h^(–1) = 20 m \ \s^(-1)` and

`t = 5` minutes `= 300 s`.

(i) From Eq. (8.5) we know that

` a = (v – u)/t`

` = ( 20 m \ \ s^(-1) – 0 m \ \s^(-1) )/(300 s)`

` = 1/(15) ms^(-2)`

(ii) From Eq. (8.7) we have

`2 a s = v^2 – u^2 = v^2 – 0`

Thus,

` s = v^2/(2a)`

` = ( (20 ms^(-1) )^2)/(2 xx (1//15) ms^(-2))`

` = 3000 m`

`= 3 km`

The acceleration of the train is ` 1/(15) m s^(-2)` and the distance travelled is `3` km.

Q 3274791656

A car accelerates uniformly from `18 km \ \ h^(–1)` to `36 km \ \ h^(–1)` in `5 s`. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.

Class 9 Chapter 8 Example 6

Class 9 Chapter 8 Example 6

We are given that

`u = 18 km \ \ h^(–1) = 5 m \ \s^(–1)`

`v = 36 km \ \ h^(–1) = 10 m \ \s^(–1)` and

`t = 5 s` .

(i) From Eq. (8.5) we have

` a = (v - u)/t`

` = ( 10 cm \ \ s^(-1) – 5 m\ \ s^(-1))/(5s)`

` = 1 m \ \ s^(–2)`

(ii) From Eq. (8.6) we have

`s = u t + 1/2 a t ^2`

` = 5 m \ \ s^(–1) xx 5 s + 1/2 xx 1 m \ \ s^(–2) xx (5 s)^2`

`= 25 m + 12.5 m`

`= 37.5 m`

The acceleration of the car is `1 m \ \ s^(–2)`

and the distance covered is `37.5 m`.

Q 3284791657

The brakes applied to a car produce an acceleration of `6 m \ \ s^(-2)` in the opposite direction to the motion. If the car takes `2 s` to stop after the application of brakes, calculate the distance it travels during this time.

Class 9 Chapter 8 Example 7

Class 9 Chapter 8 Example 7

We have been given

`a = – 6 m \ \ s^(–2) ; t = 2 s` and `v = 0 m \ \s^(–1)`.

From Eq. (8.5) we know that

`v = u + a t`

`0 = u + (– 6 m \ \s^(–2)) xx 2 s`

or `u = 12 m \ \ s^(–1)` .

From Eq. (8.6) we get

` s = u t + 1/2 a t^2`

` = (12 m s^(–1) ) xx (2 s) + 1/2 (–6 m \ \s^(–2) ) (2 s)^2`

` = 24 m – 12 m`

`= 12 m`

Thus, the car will move `12 m` before it stops after the application of brakes. Can you now appreciate why drivers are cautioned to maintain some distance between vehicles while travelling on the road?

When the velocity of an object changes, we say that the object is accelerating.

The change in the velocity could be due to change in its magnitude or the direction of the motion or both. Can you think of an example when an object does not change its magnitude of velocity but only its direction of motion?

Let us consider an example of the motion of a body along a closed path. Fig 8.9 (a) shows the path of an athlete along a rectangular track ABCD.

Let us assume that the athlete runs at a uniform speed on the straight parts AB, BC, CD and DA of the track. In order to keep himself on track, he quickly changes his speed at the corners.

How many times will the athlete have to change his direction of motion, while he completes one round? It is clear that to move in a rectangular track once, he has to change his direction of motion four times.

Now, suppose instead of a rectangular track, the athlete is running along a hexagonal shaped path ABCDEF, as shown in Fig. 8.9(b).

In this situation, the athlete will have to change his direction six times while he completes one round. What if the track was not a hexagon but a regular octagon, with eight equal sides as shown by ABCDEFGH in Fig. 8.9(c)?

It is observed that as the number of sides of the track increases the athelete has to take turns more and more often. What would happen to the shape of the track as we go on increasing the number of sides indefinitely?

If you do this you will notice that the shape of the track approaches the shape of a circle and the length of each of the sides will decrease to a point.

If the athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. The motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion.

We know that the circumference of a circle of radius `r` is given by `2πr` . If the athlete takes `t` seconds to go once around the circular path of radius r, the velocity v is given by

` v = (2 pi r)/t` (8.13)

When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.

If you carefully note, on being released the stone moves along a straight line tangential to the circular path. This is because once the stone is released, it continues to move along the direction it has been moving at that instant.

This shows that the direction of motion changed at every point when the stone was moving along the circular path.

When an athlete throws a hammer or a discus in a sports meet, he/she holds the hammer or the discus in his/her hand and gives it a circular motion by rotating his/her own body.

Once released in the desired direction, the hammer or discus moves in the direction in which it was moving at the time it was released, just like the piece of stone in the activity described above.

There are many more familiar examples of objects moving under uniform circular motion, such as the motion of the moon and the earth, a satellite in a circular orbit around the earth, a cyclist on a circular track at constant speed and so on.

The change in the velocity could be due to change in its magnitude or the direction of the motion or both. Can you think of an example when an object does not change its magnitude of velocity but only its direction of motion?

Let us consider an example of the motion of a body along a closed path. Fig 8.9 (a) shows the path of an athlete along a rectangular track ABCD.

Let us assume that the athlete runs at a uniform speed on the straight parts AB, BC, CD and DA of the track. In order to keep himself on track, he quickly changes his speed at the corners.

How many times will the athlete have to change his direction of motion, while he completes one round? It is clear that to move in a rectangular track once, he has to change his direction of motion four times.

Now, suppose instead of a rectangular track, the athlete is running along a hexagonal shaped path ABCDEF, as shown in Fig. 8.9(b).

In this situation, the athlete will have to change his direction six times while he completes one round. What if the track was not a hexagon but a regular octagon, with eight equal sides as shown by ABCDEFGH in Fig. 8.9(c)?

It is observed that as the number of sides of the track increases the athelete has to take turns more and more often. What would happen to the shape of the track as we go on increasing the number of sides indefinitely?

If you do this you will notice that the shape of the track approaches the shape of a circle and the length of each of the sides will decrease to a point.

If the athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. The motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion.

We know that the circumference of a circle of radius `r` is given by `2πr` . If the athlete takes `t` seconds to go once around the circular path of radius r, the velocity v is given by

` v = (2 pi r)/t` (8.13)

When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.

Activity _____________`8.11`

♦ Take a piece of thread and tie a small piece of stone at one of its ends. Move the stone to describe a circular path with constant speed by holding the thread at the other end, as shown in Fig. 8.10.

♦ Now, let the stone go by releasing the thread.

♦ Can you tell the direction in which the stone moves after it is released?

♦ By repeating the activity for a few times and releasing the stone at different positions of the circular path, check whether the direction in which the stone moves remains the same or not.

If you carefully note, on being released the stone moves along a straight line tangential to the circular path. This is because once the stone is released, it continues to move along the direction it has been moving at that instant.

This shows that the direction of motion changed at every point when the stone was moving along the circular path.

When an athlete throws a hammer or a discus in a sports meet, he/she holds the hammer or the discus in his/her hand and gives it a circular motion by rotating his/her own body.

Once released in the desired direction, the hammer or discus moves in the direction in which it was moving at the time it was released, just like the piece of stone in the activity described above.

There are many more familiar examples of objects moving under uniform circular motion, such as the motion of the moon and the earth, a satellite in a circular orbit around the earth, a cyclist on a circular track at constant speed and so on.