Introduction
Chapter 7 in Class 9 Science focuses on the concept of motion, which is one of the foundational concepts in physics. The chapter also delves into key mathematical tools for understanding the motion of objects, including speed, velocity, and acceleration, and provides equations of motion for solving problems related to objects in motion. In this blog post, we will focus on solving the mathematics exercises from Chapter 7 to provide you with a detailed understanding of the problems and their solutions.
Let’s break down and solve some of the typical problems in motion that you will encounter in your Class 9 exams.
Exercise 7.1 – Understanding Motion
Q1. An object travels a distance of 100 meters in 20 seconds. What is its speed?
Solution: To calculate the speed of the object, we use the formula for speed:Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}Speed=TimeDistance
Given that the distance traveled is 100 meters and the time taken is 20 seconds, we substitute these values into the formula:Speed=100 meters20 seconds=5 m/s\text{Speed} = \frac{100 \, \text{meters}}{20 \, \text{seconds}} = 5 \, \text{m/s}Speed=20seconds100meters=5m/s
Answer: The speed of the object is 5 m/s.
Q2. A car moves with a constant speed of 40 km/h for 2 hours. How much distance does it cover?
Solution: We are given:
- Speed = 40 km/h
- Time = 2 hours
To calculate the distance, we use the formula:Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}Distance=Speed×Time
Substituting the given values:Distance=40 km/h×2 hours=80 km\text{Distance} = 40 \, \text{km/h} \times 2 \, \text{hours} = 80 \, \text{km}Distance=40km/h×2hours=80km
Answer: The car covers a distance of 80 kilometers.
Exercise 7.2 – Velocity and Acceleration
Q1. A car moves with a velocity of 15 m/s and after 5 seconds its velocity becomes 25 m/s. What is its acceleration?
Solution: To find the acceleration, we use the formula:Acceleration=Change in velocityTime taken\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken}}Acceleration=Time takenChange in velocity
Given:
- Initial velocity, u=15 m/su = 15 \, \text{m/s}u=15m/s
- Final velocity, v=25 m/sv = 25 \, \text{m/s}v=25m/s
- Time taken, t=5 secondst = 5 \, \text{seconds}t=5seconds
Now, substitute the values into the formula:Acceleration=25−155=105=2 m/s2\text{Acceleration} = \frac{25 – 15}{5} = \frac{10}{5} = 2 \, \text{m/s}^2Acceleration=525−15=510=2m/s2
Answer: The acceleration of the car is 2 m/s².
Q2. A ball starts from rest and accelerates uniformly at 2 m/s22 \, \text{m/s}^22m/s2 for 10 seconds. What will be its final velocity?
Solution: We are given:
- Initial velocity, u=0 m/su = 0 \, \text{m/s}u=0m/s (since the ball starts from rest)
- Acceleration, a=2 m/s2a = 2 \, \text{m/s}^2a=2m/s2
- Time, t=10 secondst = 10 \, \text{seconds}t=10seconds
We can use the first equation of motion to find the final velocity vvv:v=u+atv = u + atv=u+at
Substituting the known values:v=0+(2×10)=20 m/sv = 0 + (2 \times 10) = 20 \, \text{m/s}v=0+(2×10)=20m/s
Answer: The final velocity of the ball will be 20 m/s.
Exercise 7.3 – Equations of Motion
Q1. A train starts from rest and accelerates uniformly at 1.5 m/s21.5 \, \text{m/s}^21.5m/s2 for 20 seconds. What distance will it cover during this time?
Solution: We are given:
- Initial velocity, u=0 m/su = 0 \, \text{m/s}u=0m/s (since the train starts from rest)
- Acceleration, a=1.5 m/s2a = 1.5 \, \text{m/s}^2a=1.5m/s2
- Time, t=20 secondst = 20 \, \text{seconds}t=20seconds
We can use the second equation of motion to calculate the distance traveled:s=ut+12at2s = ut + \frac{1}{2} a t^2s=ut+21at2
Substituting the given values:s=0+12×1.5×(20)2s = 0 + \frac{1}{2} \times 1.5 \times (20)^2s=0+21×1.5×(20)2 s=12×1.5×400=300 meterss = \frac{1}{2} \times 1.5 \times 400 = 300 \, \text{meters}s=21×1.5×400=300meters
Answer: The train will cover a distance of 300 meters.
Q2. An object moves with an initial velocity of 5 m/s5 \, \text{m/s}5m/s and a constant acceleration of 3 m/s23 \, \text{m/s}^23m/s2. Find the final velocity after 8 seconds.
Solution: We are given:
- Initial velocity, u=5 m/su = 5 \, \text{m/s}u=5m/s
- Acceleration, a=3 m/s2a = 3 \, \text{m/s}^2a=3m/s2
- Time, t=8 secondst = 8 \, \text{seconds}t=8seconds
We can use the first equation of motion to find the final velocity vvv:v=u+atv = u + atv=u+at
Substituting the given values:v=5+(3×8)=5+24=29 m/sv = 5 + (3 \times 8) = 5 + 24 = 29 \, \text{m/s}v=5+(3×8)=5+24=29m/s
Answer: The final velocity of the object will be 29 m/s.
Exercise 7.4 – Graphical Representation of Motion
Q1. The velocity-time graph of an object is a straight line with a positive slope. What can you conclude about the motion of the object?
Solution: If the velocity-time graph of an object is a straight line with a positive slope, it means that the velocity of the object is increasing at a constant rate. This is characteristic of uniform acceleration, where the object is speeding up at a constant rate over time.
In such a graph, the slope of the line represents the acceleration of the object. The area under the graph represents the displacement of the object.
Conclusion: The object is undergoing uniform acceleration.
Q2. A body moves with an initial velocity of 0 m/s and accelerates uniformly for 10 seconds. The final velocity is found to be 30 m/s. Draw the velocity-time graph for the motion and calculate the displacement.
Solution: Given:
- Initial velocity, u=0 m/su = 0 \, \text{m/s}u=0m/s
- Final velocity, v=30 m/sv = 30 \, \text{m/s}v=30m/s
- Time, t=10 secondst = 10 \, \text{seconds}t=10seconds
The acceleration can be calculated using the first equation of motion:v=u+at⇒30=0+a×10v = u + at \quad \Rightarrow \quad 30 = 0 + a \times 10v=u+at⇒30=0+a×10 a=3 m/s2a = 3 \, \text{m/s}^2a=3m/s2
Now, we know that the object accelerates uniformly, so the velocity-time graph will be a straight line with a slope of 3 m/s². The area under the graph gives the displacement, which can be calculated using:Displacement=12×(u+v)×t\text{Displacement} = \frac{1}{2} \times (u + v) \times tDisplacement=21×(u+v)×t
Substituting the known values:Displacement=12×(0+30)×10=150 meters\text{Displacement} = \frac{1}{2} \times (0 + 30) \times 10 = 150 \, \text{meters}Displacement=21×(0+30)×10=150meters
Answer: The displacement of the body is 150 meters.
Additional Important Questions for Chapter 7: Motion
1. Define motion. How is it different from rest?
- Explain the concept of motion with respect to a reference point. How is motion related to displacement?
2. An object travels a distance of 180 meters in 60 seconds. Calculate its average speed.
- Explain the method to calculate speed in both uniform and non-uniform motion.
3. A car accelerates from 20 m/s to 30 m/s in 10 seconds. Find the acceleration.
- Use the first equation of motion to calculate acceleration.
4. A ball starts from rest and accelerates uniformly for 4 seconds. If the acceleration is 3 m/s², find its final velocity.
- Use the first equation of motion to find the final velocity after 4 seconds.
5. A train accelerates from 5 m/s to 25 m/s in 10 seconds. How much distance does it cover during this time?
- Apply the second equation of motion to solve for the distance covered.
6. A car travels 100 meters in 20 seconds at a constant speed. What is the distance it will travel in 50 seconds?
- This problem involves simple speed-distance-time calculations.
7. A body is moving with a uniform speed of 10 m/s. How far will it travel in 30 minutes?
- Convert the time to seconds and calculate the total distance using the formula for speed.
8. A motorcyclist accelerates at a rate of 2 m/s² for 5 seconds. What is his final velocity if he started from rest?
- Use the first equation of motion to calculate the final velocity.
9. A stone is dropped from a height of 45 meters. How long does it take to reach the ground? (Take acceleration due to gravity g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2)
- Use the second equation of motion to find the time taken by the stone to fall.
10. A car moves with a velocity of 25 m/s for 20 seconds. How far does it travel?
- Use the equation for distance: Distance=velocity×time\text{Distance} = \text{velocity} \times \text{time}Distance=velocity×time.
11. A car accelerates uniformly from rest and covers a distance of 100 meters in 10 seconds. What is the acceleration?
- Use the second equation of motion to find acceleration.
12. A body is thrown vertically upwards with a velocity of 20 m/s. Calculate the time taken to reach the maximum height and the maximum height attained.
- Use the equations of motion to solve for time and height. Assume g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2.
13. A car starts from rest and accelerates at a rate of 4 m/s24 \, \text{m/s}^24m/s2 for 6 seconds. What is its final speed?
- Use the first equation of motion to calculate the final speed.
14. A vehicle moves with a constant speed of 72 km/h. What is its speed in meters per second (m/s)?
- Convert the given speed into m/s.
15. A bullet is fired with an initial velocity of 100 m/s. After 2 seconds, its velocity is reduced to 80 m/s. Find its acceleration.
- Use the formula for acceleration: a=Δvta = \frac{\Delta v}{t}a=tΔv.
16. A body moves with an initial velocity of 4 m/s and an acceleration of 2 m/s22 \, \text{m/s}^22m/s2. Calculate the velocity after 5 seconds.
- Apply the first equation of motion.
17. A runner covers a distance of 400 meters in 50 seconds. Calculate the runner’s speed and velocity.
- Find the speed first, then discuss the concept of velocity in this case.
18. What does a distance-time graph of an object moving with a constant velocity look like?
- Describe the shape and slope of the graph.
19. A ball is thrown vertically upward with an initial velocity of 10 m/s. What will be its velocity after 1 second? (Take g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)
- Use the first equation of motion to solve for the velocity after 1 second.
20. An object moves with uniform acceleration. If its initial velocity is 0 m/s and it covers a distance of 80 meters in 8 seconds, calculate its acceleration.
- Use the second equation of motion to calculate acceleration.
Concepts Tested by These Questions
These 20 questions cover a broad range of topics from Chapter 7 – Motion, ensuring that you understand and are capable of solving problems related to:
- Speed and velocity: Understanding the differences and calculating both in various scenarios.
- Acceleration: How to calculate acceleration from initial and final velocities or through graphical representation.
- Equations of motion: Applying the first, second, and third equations of motion to solve real-life problems.
- Graphical representation: Analyzing and interpreting graphs, such as distance-time and velocity-time graphs.
- Free fall and gravitational acceleration: Solving problems related to objects in free fall under the influence of gravity.
Practicing these important questions will help reinforce the core concepts of motion, which are fundamental to understanding classical mechanics. Regular practice with a variety of questions, including real-life applications and theoretical problems, will ensure a strong grasp of the chapter.
Good luck with your preparation! Stay tuned for more practice materials and tips
Also Read: Understanding Motion: A Detailed Guide to Class 9 Science Chapter 7
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