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Chapter Analysis
Intermediate18 pages • EnglishQuick Summary
This chapter introduces the concept of motion, which encompasses both the physical movement of objects and the mathematical context, such as speed, velocity, and acceleration. The chapter discusses how motion can be represented graphically through distance-time and velocity-time graphs. Equations of motion are derived for uniformly accelerated linear motion, and concepts like uniform circular motion are also explored, providing a foundational understanding of how motion is described and analyzed.
Key Topics
- •Motion and its types
- •Speed and velocity
- •Uniform and non-uniform motion
- •Distance-time and velocity-time graphs
- •Equations of motion
- •Uniform circular motion
- •Acceleration
Learning Objectives
- ✓Understand the concept of motion along a straight line
- ✓Learn to differentiate between speed and velocity
- ✓Gain the ability to interpret distance-time and velocity-time graphs
- ✓Derive and apply equations of motion for uniformly accelerated linear motion
- ✓Explore concepts of circular motion and acceleration
Questions in Chapter
Distinguish between speed and velocity.
Page 74
Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Page 74
What does the odometer of an automobile measure?
Page 74
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?
Page 85
A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km h–1 in 10 minutes. Find its acceleration.
Page 78
Additional Practice Questions
A bullet train moves from rest to a speed of 300 km/h in 5 minutes. Calculate the acceleration.
mediumAnswer: To calculate the acceleration, we first convert the speed to m/s: 300 km/h = 300,000 m/3600 s = 83.33 m/s. The train moves from rest (0 m/s) to 83.33 m/s in 300 s. Therefore, acceleration a = Δv/Δt = 83.33 m/s / 300 s = 0.278 m/s².
If a cyclist moves in a straight line with constant speed of 15 m/s, how long will it take to cover a distance of 1.5 km?
easyAnswer: Given the speed v = 15 m/s and distance s = 1.5 km = 1500 m. Time t = s/v = 1500 m / 15 m/s = 100 s.
Describe how velocity-time graphs can be used to calculate distance traveled.
mediumAnswer: The area under a velocity-time graph gives the distance traveled. If the graph is a straight line, the area is a rectangle; for non-uniform motion with varying velocity, the shape can be more complex, such as a triangle or trapezoid.
Derive the relationship between displacement, initial velocity, acceleration, and time for an object moving with uniform acceleration.
hardAnswer: Start from the basic equations of motion: v = u + at and s = ut + 1/2 at². Solving these simultaneously provides the relationship: s = (v² - u²)/2a.
Calculate the time it takes for a stone dropped from a height of 100 m to hit the ground. Assume g = 9.8 m/s², and neglect air resistance.
mediumAnswer: Use the equation s = ut + 1/2 at², where u = 0, s = 100 m, and a = g = 9.8 m/s². 100 = 0 + 1/2 * 9.8 * t², solving gives t ≈ 4.52 s.
NCERT Exemplar
AvailablePractice with NCERT Exemplar problems and solutions for this chapter. Exemplar problems are designed to help you master the concepts with advanced-level questions.
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