In the previous section, we learned about Displacement. In this section we will see Uniform and Non-uniform Motion.
In the next section, we will learn about Rate of change of velocity.
Uniform motion and non uniform motion
■ Consider an object moving in a straight line. Let it travel:
• 5 m in the first second.
• 5 more metre in the next second,
• 5 m in the third second and
• 5 m in the fourth second.
So the distance covered by the object in every second is equal. It is 5 m
■ If an object covers equal distances in equal intervals of time, it is said to be in uniform motion.
• In our day to day life, such uniform motion is occurs only rarely. What we encounter most is non-uniform motion.
• For example, a car travelling in a crowded street will not be able to maintain steady speed.
♦ It may be able to cover 25 m in the first 5 seconds.
♦ If the crowd increases after those five seconds, the speed will have to be so decreased that, it travels only 5 m in the next 5 seconds.
• The motion of the car is not uniform in this case.
• If another object is moving slowly, it will cover a smaller distance within the same time ‘t’
• So we need a method to compare the motions of different objects.
• For that comparison, we use a unit called ‘metre per second’. In short form it is written as m/s or ms-1
• For greater speeds we use kilometres per hour. In short form it is written as km/hr or kmph
■ If some one tells us that "a car is moving at 40 km/hr", we can do the following simple analysis:
1. We see that car moving (see fig.1.4 below)
2. We make a 'mark 1' on the ground.
3. At the instant when the car crosses that mark, we start a stop watch
4. Exactly after 1 hour, we check the position of the car.
5. The position will be 40 km away from the first mark. This is shown in the fig.1.4 below:
6. That means, the car has travelled 40 km in one hour.
• That is 40 km per hour.
• So the speed is 40 km/hr.
7. If after 1 more hour, we check the position of the car, it will be 80 km away from Mark 1.
Eq.1.1:
Average speed = Total distance⁄Total time
Let us see an example:
• An object travels 16 m in 4 seconds. Then it travels another 16 m in 2 seconds.
• So the total distance travelled = 16 + 16 = 32 m
• Total time taken for this travel = 6 seconds
• Thus average speed = Total distance⁄Total time = 32 m⁄6 s = 5.33 m/s
Examples:
• An object is moving with a velocity of 20 m/s along a line which makes an angle 60o with the y axis. See fig.1.5(a)
• A car is moving with a velocity of 35 km/hr in the North east direction. See fig.1.5(b)
Change in velocity
We have seen that velocity has both magnitude and direction. So a velocity can be changed by any one of the following three methods:
• Changing speed keeping direction the same
♦ An example: An object was travelling with a velocity of 20 m/s, at an angle of 25o with the x-axis. After some time, the speed changed to 15 m/s. In this case, the velocity has changed even if the object continues to travel in the same direction. The new velocity should be specified by the new speed and the same direction
• Changing direction keeping speed the same
♦ An example: A car was travelling with a velocity of 30 Kmph in the north east direction. After some time, the car changed course. In this case, the velocity has changed even if the car continues to travel with the same speed. The new velocity should be specified by the speed 30 kmph and new direction.
• Changing both speed and direction
♦ An example: A car was travelling with a velocity of 30 Kmph in the north east direction. After some time, the car changed course. The speed also changed to 20 kmph. In this case, the velocity has changed due to change in both speed and direction.
■ In our day to day life, we are able to travel along straight roads with no bends or curves. But it is difficult to travel with a uniform speed even if the road is straight. That means, we are travelling with non uniform velocity. In such cases, how will we express the velocity?
Ans: If the velocity is not uniform, we will have to use ‘average velocity’.
It is calculated as follows:
Eq.1.2:
Average velocity = Total displacement⁄Total time
This calculation can be demonstrated with an example:
1. Consider fig.1.6 below. A car is initially at rest at the origin point 'O'.
2. From O, it travels in the positive direction of the x-axis for a distance of 4 Km, and reaches 'A'. The journey from O to A takes 10 minutes.
■ Average speed = Total distance⁄Total time = 400 Km⁄8 h = 400000 m⁄28800 s = 13.88 m s-1
Solved example 1.2
Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Solution:
To find average speed:
• So average speed = 180 m⁄1 min = 180 m⁄60 s = 3 m s-1
To find average velocity:
• 5 m in the first second.
• 5 more metre in the next second,
• 5 m in the third second and
• 5 m in the fourth second.
So the distance covered by the object in every second is equal. It is 5 m
■ If an object covers equal distances in equal intervals of time, it is said to be in uniform motion.
• In our day to day life, such uniform motion is occurs only rarely. What we encounter most is non-uniform motion.
• For example, a car travelling in a crowded street will not be able to maintain steady speed.
♦ It may be able to cover 25 m in the first 5 seconds.
♦ If the crowd increases after those five seconds, the speed will have to be so decreased that, it travels only 5 m in the next 5 seconds.
• The motion of the car is not uniform in this case.
Speed
• If an object is moving fast, it will cover a larger distance within a given time ‘t’• If another object is moving slowly, it will cover a smaller distance within the same time ‘t’
• So we need a method to compare the motions of different objects.
• For that comparison, we use a unit called ‘metre per second’. In short form it is written as m/s or ms-1
• For greater speeds we use kilometres per hour. In short form it is written as km/hr or kmph
■ If some one tells us that "a car is moving at 40 km/hr", we can do the following simple analysis:
1. We see that car moving (see fig.1.4 below)
2. We make a 'mark 1' on the ground.
3. At the instant when the car crosses that mark, we start a stop watch
4. Exactly after 1 hour, we check the position of the car.
5. The position will be 40 km away from the first mark. This is shown in the fig.1.4 below:
 |
| Fig.1.4 |
• That is 40 km per hour.
• So the speed is 40 km/hr.
7. If after 1 more hour, we check the position of the car, it will be 80 km away from Mark 1.
So we saw how to express the speed of an object. The car we saw was in an uniform motion. That is., it covers equal distances in equal intervals of time. What if the car was moving in a crowded street? Then it’s motion will not be uniform. In such a case, how will we express the speed?
For non uniform motion, we will have to obtain average speed. It is given by:Eq.1.1:
Average speed = Total distance⁄Total time
Let us see an example:
• An object travels 16 m in 4 seconds. Then it travels another 16 m in 2 seconds.
• So the total distance travelled = 16 + 16 = 32 m
• Total time taken for this travel = 6 seconds
• Thus average speed = Total distance⁄Total time = 32 m⁄6 s = 5.33 m/s
Velocity
■ Velocity is obtained when we specify direction also with speed.Examples:
• An object is moving with a velocity of 20 m/s along a line which makes an angle 60o with the y axis. See fig.1.5(a)
• A car is moving with a velocity of 35 km/hr in the North east direction. See fig.1.5(b)
 |
| Fig.1.5 |
We have seen that velocity has both magnitude and direction. So a velocity can be changed by any one of the following three methods:
• Changing speed keeping direction the same
♦ An example: An object was travelling with a velocity of 20 m/s, at an angle of 25o with the x-axis. After some time, the speed changed to 15 m/s. In this case, the velocity has changed even if the object continues to travel in the same direction. The new velocity should be specified by the new speed and the same direction
• Changing direction keeping speed the same
♦ An example: A car was travelling with a velocity of 30 Kmph in the north east direction. After some time, the car changed course. In this case, the velocity has changed even if the car continues to travel with the same speed. The new velocity should be specified by the speed 30 kmph and new direction.
• Changing both speed and direction
♦ An example: A car was travelling with a velocity of 30 Kmph in the north east direction. After some time, the car changed course. The speed also changed to 20 kmph. In this case, the velocity has changed due to change in both speed and direction.
• Consider an object travelling along a straight line. This line is ‘straight’, with no bends or curves.
• So that object is travelling with absolutely no change in direction.
• If that object is travelling with an uniform speed, it is indeed travelling with an uniform velocity.
■ In our day to day life, we are able to travel along straight roads with no bends or curves. But it is difficult to travel with a uniform speed even if the road is straight. That means, we are travelling with non uniform velocity. In such cases, how will we express the velocity?
Ans: If the velocity is not uniform, we will have to use ‘average velocity’.
It is calculated as follows:
Eq.1.2:
Average velocity = Total displacement⁄Total time
This calculation can be demonstrated with an example:
1. Consider fig.1.6 below. A car is initially at rest at the origin point 'O'.
 |
| Fig.1.6 |
3. Then it travels in the positive direction of the y-axis for a distance of 3 Km, and reaches 'B'. The journey from A to B takes 5 minutes.
4. B is the final position of the car. So displacement of the car is OB = 5 Km
5. Total time for the travel = 10 + 5 = 15 minutes
6. Thus average velocity = Total displacement⁄Total time = 5 Km⁄15 mins
• 15 mins = 0.25 = 1/4 hr
• So average velocity = 5 Km⁄15 mins = 5 Km⁄1/4 hr = 20 kmph
• 8 h = 8 × 60 × 60 = 28800 seconds4. B is the final position of the car. So displacement of the car is OB = 5 Km
5. Total time for the travel = 10 + 5 = 15 minutes
6. Thus average velocity = Total displacement⁄Total time = 5 Km⁄15 mins
• 15 mins = 0.25 = 1/4 hr
• So average velocity = 5 Km⁄15 mins = 5 Km⁄1/4 hr = 20 kmph
We will now see some solved examples
Solved example 1.1
The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h-1 and m s-1.
Solution:
• Initial reading = 2000 Km
• Final reading = 2400 Km
• So distance travelled = 2400 – 2000 = 400 Km
• Total time = 8 h
We have:
■ Average speed = Total distance⁄Total time = 400 Km⁄8 h = 50 km h-1
• 400 Km = 400 × 1000 = 400000 m■ Average speed = Total distance⁄Total time = 400 Km⁄8 h = 400000 m⁄28800 s = 13.88 m s-1
Solved example 1.2
Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Solution:
To find average speed:
• Distance travelled = 180 m
• Total time = 1 m
We have:
■ Average speed = Total distance⁄Total time = 180 m⁄1 m
• 1 mins = 60 s• So average speed = 180 m⁄1 min = 180 m⁄60 s = 3 m s-1
To find average velocity:
• Displacement = 0 m (∵ initial position and final position are the same)
• Total time = 1 m
We have:
■ Average velocity = Total displacement⁄Total time = 0 m⁄1 m = 0 m/min = 0 m s-1
In the next section, we will learn about Rate of change of velocity.
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