In the previous section, we completed the discussion on the motion of objects. In this section we will see some solved examples.
Solved example 1.16
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution:
1. Diameter of the circular track = d = 200 m. So radius = r = 100 m
2. Perimeter of the track = 2πr = πd = 200π m
3. Time for completing one round = t = 40 s
4. Speed of the athlete = distance⁄time = πd⁄t = 200π⁄40 = 5π m s-1.
5. So the athlete covers a distance of 5π m in 1 s. We have to find the distance that the athlete covers in 2 mins and 20 s
6. 2 mins 20 s = 140 s
7. So the distance covered in this time = 140 × 5π = 700π m
8. Now we have to find the displacement at the end of 140 s
9. When the athlete covers one perimeter 200π, from the starting point, he reaches back to his starting point. At that time, his displacement is zero
10. In this way, when he travels 700π, he reaches back the starting point 3 times. After the third time he travels a 'certain distance' to make the 700π. This 'certain distance is the displacement'. It can be calculated as follows:
11. 700π = 3 × 200π + 100π.
So the 'certain distance' = displacement = 100π.
Solved example 1.17
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Solution:
Case 1. From A to B:
1. Distance = 300 m
2. Time = 2 mins 30 s = 150 s
3. Speed = distance⁄time = 300⁄150 = 2 m s-1.
4. Velocity = displacement⁄time
• Here displacement = 300 m
• So velocity = displacement⁄time = 300⁄150 = 2 m s-1.
Case 2. From A to C
1. Distance = (300 m + 100 m) = 400 m
2. Time = (2 mins 30 s + 1 min) = 3 mins 30 s = 210 s
3. Speed = distance⁄time = 400⁄210 = 1.904 m s-1.
4. Velocity = displacement⁄time
• Here displacement = 200 m
• So velocity = displacement⁄time = 200⁄210 = 0.952 m s-1.
Solved example 1.18
Abdul, while driving to school, computes the average speed for his trip to be 20 kmph. On his return trip along the same route, there is less traffic and the average speed is 30 kmph. What is the average speed for Abdul’s trip?
Solution:
1. Let the distance from home to school be 's'
2. Let the time required in the morning be t1
3. Then Abdul calculated his average speed in the morning to be 20 kmph in the following way:
Speed = distance⁄time = s⁄t1 = 20 kmph
4. From this we get: t1 = s⁄20 .
5. In the evening he travels the same distance 's'
6. Let the time required in the evening be t2
7. Then he calculated his average speed in the evening to be 30 kmph in the following way:
Speed = distance⁄time = s⁄t2 = 30 kmph
8. From this we get: t2 = s⁄30 .
9. So total time required for the travel = (s⁄20 + s⁄30) = 5s⁄60.
10. Total distance = s + s = 2s
11. So average speed = total distance⁄total time = 2s ÷ 5s⁄60 = 2s × 60⁄5s = 24 kmph
Solved example 1.19
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-2 for 8.0 s. How far does the boat travel during this time?
Solution:
1. The motorboat starts from rest. So initial velocity u = 0
• Time of travel = 8 s
• Acceleration = 3 m s-2.
2. We want the distance s
The second equation of motion connects the above quantities. So we will use it:
3. s = ut + 1⁄2 at2 ⇒ s = 0 × 300 + 1⁄2 × 3 × 82 ⇒ s = 96 m
Solved example 1.20
A driver of a car travelling at 52 kmph applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 30 kmph in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Solution:
The fig. shows the speed-time graph for the two cars:
1. The yellow line AB is the speed-time graph of the first car. The purple line PQ is the speed-time graph of the second car.
Note that, the speeds are converted from kmph to m s-1.
2. The area enclosed by a speed-time graph and the time axis will give the distance travelled by the object. We saw the details here. The areas are shown shaded in the above fig.
3. First car:
Area = Area of the triangle OAB = 1⁄2 × 5 × 14.44 = 36.1 m
[ ∵ Base = 5 units and altitude = 14.44 units]
4. Second car :
Area = Area of the triangle OPQ = 1⁄2 × 10 × 8.33 = 41.65 m
[ ∵ Base = 10 units and altitude = 8.33 units]
Solved example 1.21
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?
Solution:
1. Height of fall = distance of travel = s = 20 m
2. Acceleration = a = 10 m s-2.
3. The ball falls from rest. So it has an initial velocity u = 0
4. We have to find the final velocity v. We can use the third equation of motion. Because it connects all these quantities
5. So we can write:
v2 = u2 + 2as ⇒ v2 = 02 + 2 × 10 × 20 ⇒ v2 = 400 ⇒ v = 20 m s-1.
6. To find the time, we can use the first equation of motion:
v = u + at ⇒ 20 = 0 + 10t ⇒ 20 = 10 t ⇒ t = 2 s
Solved example 1.22
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
1. Radius of the orbit = r = 42250 km
2. Perimeter of the orbit = 2πr = 2 × π × 42250 = 84500π km = 84500000 m
3. Time for completing one round revolution = t = 24 hours = 24 × 60 × 60 = 86400 s
4. Speed of the satellite = distance⁄time = 84500000π⁄86400 = 3070.95 m s-1
So we have completed the discussion on the 'Motion of objects'. In the next chapter, we will see 'Forces acting on objects'.
Solved example 1.16
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution:
1. Diameter of the circular track = d = 200 m. So radius = r = 100 m
2. Perimeter of the track = 2πr = πd = 200π m
3. Time for completing one round = t = 40 s
4. Speed of the athlete = distance⁄time = πd⁄t = 200π⁄40 = 5π m s-1.
5. So the athlete covers a distance of 5π m in 1 s. We have to find the distance that the athlete covers in 2 mins and 20 s
6. 2 mins 20 s = 140 s
7. So the distance covered in this time = 140 × 5π = 700π m
8. Now we have to find the displacement at the end of 140 s
9. When the athlete covers one perimeter 200π, from the starting point, he reaches back to his starting point. At that time, his displacement is zero
10. In this way, when he travels 700π, he reaches back the starting point 3 times. After the third time he travels a 'certain distance' to make the 700π. This 'certain distance is the displacement'. It can be calculated as follows:
11. 700π = 3 × 200π + 100π.
So the 'certain distance' = displacement = 100π.
Solved example 1.17
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Solution:
Case 1. From A to B:
1. Distance = 300 m
2. Time = 2 mins 30 s = 150 s
3. Speed = distance⁄time = 300⁄150 = 2 m s-1.
4. Velocity = displacement⁄time
• Here displacement = 300 m
• So velocity = displacement⁄time = 300⁄150 = 2 m s-1.
Case 2. From A to C
1. Distance = (300 m + 100 m) = 400 m
2. Time = (2 mins 30 s + 1 min) = 3 mins 30 s = 210 s
3. Speed = distance⁄time = 400⁄210 = 1.904 m s-1.
4. Velocity = displacement⁄time
• Here displacement = 200 m
• So velocity = displacement⁄time = 200⁄210 = 0.952 m s-1.
Solved example 1.18
Abdul, while driving to school, computes the average speed for his trip to be 20 kmph. On his return trip along the same route, there is less traffic and the average speed is 30 kmph. What is the average speed for Abdul’s trip?
Solution:
1. Let the distance from home to school be 's'
2. Let the time required in the morning be t1
3. Then Abdul calculated his average speed in the morning to be 20 kmph in the following way:
Speed = distance⁄time = s⁄t1 = 20 kmph
4. From this we get: t1 = s⁄20 .
5. In the evening he travels the same distance 's'
6. Let the time required in the evening be t2
7. Then he calculated his average speed in the evening to be 30 kmph in the following way:
Speed = distance⁄time = s⁄t2 = 30 kmph
8. From this we get: t2 = s⁄30 .
9. So total time required for the travel = (s⁄20 + s⁄30) = 5s⁄60.
10. Total distance = s + s = 2s
11. So average speed = total distance⁄total time = 2s ÷ 5s⁄60 = 2s × 60⁄5s = 24 kmph
Solved example 1.19
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-2 for 8.0 s. How far does the boat travel during this time?
Solution:
1. The motorboat starts from rest. So initial velocity u = 0
• Time of travel = 8 s
• Acceleration = 3 m s-2.
2. We want the distance s
The second equation of motion connects the above quantities. So we will use it:
3. s = ut + 1⁄2 at2 ⇒ s = 0 × 300 + 1⁄2 × 3 × 82 ⇒ s = 96 m
Solved example 1.20
A driver of a car travelling at 52 kmph applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 30 kmph in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Solution:
The fig. shows the speed-time graph for the two cars:
1. The yellow line AB is the speed-time graph of the first car. The purple line PQ is the speed-time graph of the second car.
Note that, the speeds are converted from kmph to m s-1.
2. The area enclosed by a speed-time graph and the time axis will give the distance travelled by the object. We saw the details here. The areas are shown shaded in the above fig.
3. First car:
Area = Area of the triangle OAB = 1⁄2 × 5 × 14.44 = 36.1 m
[ ∵ Base = 5 units and altitude = 14.44 units]
4. Second car :
Area = Area of the triangle OPQ = 1⁄2 × 10 × 8.33 = 41.65 m
[ ∵ Base = 10 units and altitude = 8.33 units]
Solved example 1.21
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?
Solution:
1. Height of fall = distance of travel = s = 20 m
2. Acceleration = a = 10 m s-2.
3. The ball falls from rest. So it has an initial velocity u = 0
4. We have to find the final velocity v. We can use the third equation of motion. Because it connects all these quantities
5. So we can write:
v2 = u2 + 2as ⇒ v2 = 02 + 2 × 10 × 20 ⇒ v2 = 400 ⇒ v = 20 m s-1.
6. To find the time, we can use the first equation of motion:
v = u + at ⇒ 20 = 0 + 10t ⇒ 20 = 10 t ⇒ t = 2 s
Solved example 1.22
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
1. Radius of the orbit = r = 42250 km
2. Perimeter of the orbit = 2πr = 2 × π × 42250 = 84500π km = 84500000 m
3. Time for completing one round revolution = t = 24 hours = 24 × 60 × 60 = 86400 s
4. Speed of the satellite = distance⁄time = 84500000π⁄86400 = 3070.95 m s-1
So we have completed the discussion on the 'Motion of objects'. In the next chapter, we will see 'Forces acting on objects'.
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