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Chapter Analysis
Intermediate25 pages • EnglishQuick Summary
This chapter delves into the concepts of motion in a straight line, focusing on understanding kinematic equations for uniformly accelerated motion. It explores instantaneous velocity, speed, and acceleration, as well as relative velocity. The content provides a foundation for describing motion without delving into its causes, laying the groundwork for further exploration in subsequent chapters.
Key Topics
- •Motion in a straight line
- •Instantaneous velocity and speed
- •Acceleration
- •Kinematic equations for uniformly accelerated motion
- •Relative velocity
- •Velocity-time graphs
- •Displacement and distance
- •Instantaneous acceleration
Learning Objectives
- ✓Understand and describe motion using velocity and acceleration.
- ✓Apply kinematic equations to solve motion problems.
- ✓Differentiate between instantaneous and average values of velocity and acceleration.
- ✓Interpret and analyze motion graphs for different types of acceleration.
- ✓Explain the concept of relative velocity and apply it to solve related problems.
Questions in Chapter
In which of the following examples of motion, can the body be considered approximately a point object: (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table.
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The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below ; (a) (A/B) lives closer to the school than (B/A) (b) (A/B) starts from the school earlier than (B/A) (c) (A/B) walks faster than (B/A) (d) A and B reach home at the (same/different) time (e) (A/B) overtakes (B/A) on the road (once/twice).
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A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
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A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
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A player throws a ball upwards with an initial speed of 29.4 m s–1. (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g = 9.8 m s–2 and neglect air resistance).
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Additional Practice Questions
If an object is moving with a velocity of 10 m/s and accelerates at 2 m/s², what will be its velocity after 5 seconds?
easyAnswer: The final velocity can be calculated using the equation v = u + at. Here, u = 10 m/s, a = 2 m/s², and t = 5 s. So, v = 10 + (2 * 5) = 20 m/s.
A car accelerates uniformly from rest to a speed of 30 m/s in 10 seconds. What is the distance covered by the car?
mediumAnswer: The distance covered can be calculated using the equation s = ut + 0.5at². Here, u = 0 m/s (rest), v = 30 m/s, t = 10 s. The acceleration a can be obtained using the formula v = u + at, so a = (v-u)/t = 3 m/s². Now, s = (0 * 10) + 0.5 * 3 * (10)² = 150 m.
An object moves with a constant velocity of 15 m/s for 8 s. Calculate the distance travelled by the object.
easyAnswer: Distance travelled can be calculated using the formula distance = velocity * time. Here, distance = 15 m/s * 8 s = 120 m.
A particle moves along a straight line with an initial velocity of 5 m/s and a constant acceleration of 4 m/s². What is its displacement after 3 s?
mediumAnswer: The displacement can be calculated using the equation s = ut + 0.5at². Here, u = 5 m/s, a = 4 m/s², and t = 3 s. So, s = (5 * 3) + 0.5 * 4 * (3)² = 15 + 18 = 33 m.
A train is decelerating at 1 m/s² as it comes to a stop. If it comes to a stop in 20 seconds, what was its initial velocity?
mediumAnswer: The initial velocity can be calculated using the equation v = u + at. Here, v = 0 (since the train stops), a = -1 m/s² (deceleration), and t = 20 s. Solving 0 = u + (-1)*20 gives u = 20 m/s.