Motion in a Straight Line

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Summary

Summary of Motion in a Straight Line

Introduction to Motion: Motion is the change in position of an object over time. It can be observed in various forms such as walking, running, or even the movement of air.
Key Concepts:
- Instantaneous Velocity: The limit of average velocity as the time interval approaches zero.
- Acceleration: The rate of change of velocity over time.
- Kinematic Equations: Equations that relate displacement, time, initial velocity, final velocity, and acceleration for uniformly accelerated motion.
- Relative Velocity: The velocity of an object as observed from another moving object.
Kinematic Equations:
- For uniformly accelerated motion:
  - Equation 1: v = v₀ + at
  - Equation 2: x = v₀t + 1/2 at²
  - Equation 3: v² = v₀² + 2ax
Average Speed vs. Average Velocity: Average speed is always greater than or equal to the magnitude of average velocity over a given time interval.
Acceleration Types:
- Average Acceleration: Change in velocity divided by the time interval.
- Instantaneous Acceleration: The limit of average acceleration as the time interval approaches zero.
Graphical Representations:
- Position-time graphs show the position of an object over time.
- Velocity-time graphs illustrate how velocity changes over time.
- Acceleration-time graphs depict changes in acceleration over time.
Important Points to Ponder:
- The choice of origin and direction affects the signs of displacement, velocity, and acceleration.
- Zero velocity does not imply zero acceleration.
- The kinematic equations apply only under constant acceleration conditions.

Learning Objectives

Understand the concept of motion and its significance in the universe.
Define and differentiate between instantaneous velocity and average velocity.
Explain the concept of acceleration and its types (average and instantaneous).
Apply kinematic equations for uniformly accelerated motion to solve problems.
Analyze relative velocity and its implications in different frames of reference.
Interpret and construct position-time and velocity-time graphs for various types of motion.
Recognize the importance of choosing a reference point and direction in motion analysis.

Detailed Notes

Chapter 2: Motion in a Straight Line

2.1 Introduction

Motion is common to everything in the universe.
Describes how objects change position over time.
Focus on rectilinear motion (motion along a straight line).
Key concepts: velocity, acceleration, and relative velocity.

2.2 Instantaneous Velocity and Speed

Instantaneous velocity is defined as the limit of average velocity as the time interval approaches zero.
Velocity at a specific instant equals the slope of the tangent on the position-time graph.

2.3 Acceleration

Average acceleration is the change in velocity divided by the time interval.
Instantaneous acceleration is defined as the limit of average acceleration as the time interval approaches zero.
The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant.

2.4 Kinematic Equations for Uniformly Accelerated Motion

For uniformly accelerated rectilinear motion, the following kinematic equations apply:
- Equation 1: v = v₀ + at
- Equation 2: x = v₀t + (1/2)at²
- Equation 3: v² = v₀² + 2ax
Where:
- v = final velocity
- v₀ = initial velocity
- a = acceleration
- x = displacement
- t = time

2.5 Relative Velocity

Introduces the concept of relative velocity to understand motion in different frames of reference.

Points to Ponder

The choice of origin and positive direction affects the signs of displacement, velocity, and acceleration.
Acceleration direction relative to velocity indicates whether speed is increasing or decreasing.
Zero velocity does not imply zero acceleration; a particle can be at rest while having non-zero acceleration.
Kinematic equations are applicable for one-dimensional motion with constant acceleration, considering proper signs for quantities.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Neglecting the size of objects: When treating objects as point-like, ensure their size is much smaller than the distance they move in a reasonable time.
Misunderstanding acceleration: Remember that a particle can have zero speed but non-zero acceleration at an instant (e.g., at the highest point of a thrown ball).
Confusing average and instantaneous quantities: Average speed is not always equal to instantaneous speed; be clear about the definitions.
Incorrect sign conventions: The choice of positive direction affects the signs of displacement, velocity, and acceleration. Always specify your choice before solving problems.
Assuming constant acceleration: Kinematic equations apply only when acceleration is constant; check the conditions before using them.

Tips for Exam Preparation

Practice with graphs: Familiarize yourself with interpreting position-time and velocity-time graphs, as they are crucial for understanding motion.
Work through examples: Solve various problems involving different types of motion to solidify your understanding of concepts like velocity, acceleration, and kinematic equations.
Review definitions: Ensure you know the definitions of key terms such as instantaneous velocity, average velocity, and acceleration, including their units and dimensions.
Understand the relationship between quantities: Be clear on how displacement, velocity, and acceleration relate to each other, especially in uniformly accelerated motion.

Practice & Assessment

Multiple Choice Questions

A railway carriage moving without jerks between two stations.

A monkey sitting on top of a man cycling smoothly on a circular track.

A spinning cricket ball that turns sharply on hitting the ground.

A tumbling beaker that has slipped off the edge of a table.

Correct Answer: A

Solution:

A railway carriage moving without jerks can be considered a point object because its size is much smaller compared to the distance it travels.

10.2 m

20.4 m

30.6 m

40.8 m

Correct Answer: B

Solution:

Using the equation

v^2 = u^2 - 2gh

, where

v = 0

at the maximum height,

u = 20

m/s, and

g = 9.8

m/s², we solve for

h

0 = 20^2 - 2 \times 9.8 \times h

, giving

h = 20.4

108 m/s

42 m/s

150 m/s

192 m/s

Correct Answer: A

Solution:

Convert the speeds to m/s: Police car speed = 30 km/h = 8.33 m/s, Thief's car speed = 192 km/h = 53.33 m/s. Relative speed of bullet = Bullet speed - Thief's car speed = 150 - 53.33 = 96.67 m/s.

17 seconds

19 seconds

21 seconds

23 seconds

Correct Answer: B

Solution:

The drunkard effectively covers 2 meters (5 forward - 3 backward) in 8 seconds (5+3 steps). To cover 12 meters, he needs 4 such cycles, taking

4 \times 8 = 32

seconds. In the next cycle, he will cover 5 meters forward in 5 seconds, reaching 17 meters, which means he falls into the pit after a total of

32 + 5 = 37

seconds.

16 seconds

20 seconds

24 seconds

28 seconds

Correct Answer: C

Solution:

In every 8 seconds, the drunkard moves 2 meters forward (5 steps forward and 3 steps backward). To cover 13 meters, he needs 5 complete cycles (40 seconds) and an additional 3 steps forward (3 seconds), totaling 24 seconds.

A railway carriage moving without jerks between two stations.

A monkey sitting on top of a man cycling smoothly on a circular track.

A spinning cricket ball that turns sharply on hitting the ground.

A tumbling beaker that has slipped off the edge of a table.

Correct Answer: A

Solution:

A railway carriage moving without jerks between two stations can be considered a point object because its size is negligible compared to the distance it travels.

36.0 m/s

40.5 m/s

45.0 m/s

49.5 m/s

Correct Answer: B

Solution:

The speed of the ball just before it hits the floor for the first time is found using

v^2 = u^2 + 2gh

, where

u = 0

g = 9.8 \text{ m/s}^2

, and

h = 90 \text{ m}

. Solving gives

v = \sqrt{2 \times 9.8 \times 90} = 42 \text{ m/s}

. After losing one tenth of its speed, the speed is

42 \times 0.9 = 37.8 \text{ m/s}

. When the ball rebounds, it reaches a height where its speed again becomes

0

, and it falls back gaining speed. The speed just before hitting the floor for the second time is

37.8 \times 1.1 = 40.5 \text{ m/s}

25 m

35 m

45 m

15 m

Correct Answer: A

Solution:

Displacement is the integral of velocity over time.

x(t) = \int v(t) \, dt = \int (4t - 3) \, dt = 2t^2 - 3t + C

. From

t = 0

t = 5

x(5) - x(0) = [2(5)^2 - 3(5)] - [2(0)^2 - 3(0)] = 50 - 15 = 25 \text{ m}

Upward

Downward

Horizontal

Zero

Correct Answer: B

Solution:

The acceleration due to gravity is always downward, even when the ball is moving upwards.

5.83 km/h

6.25 km/h

6.75 km/h

7.00 km/h

Correct Answer: A

Solution:

The total distance traveled is

2.5 \text{ km} \times 2 = 5 \text{ km}

. The time taken for the onward journey is

\frac{2.5}{5} = 0.5 \text{ h}

, and for the return journey is

\frac{2.5}{7.5} = \frac{1}{3} \text{ h}

. The total time taken is

0.5 + \frac{1}{3} = \frac{5}{6} \text{ h}

. The average speed is

\frac{5}{\frac{5}{6}} = 6 \text{ km/h}

42 m/s

39.6 m/s

44.3 m/s

36 m/s

Correct Answer: B

Solution:

Using the equation of motion

v^2 = u^2 + 2as

, where

u = 0

a = g = 9.8 \text{ m/s}^2

, and

s = 90 \text{ m}

, we find

v = \sqrt{2 \times 9.8 \times 90} = 39.6 \text{ m/s}

A straight line parallel to the time axis on a velocity-time graph.

A parabola on a position-time graph.

A horizontal line on a position-time graph.

A sinusoidal wave on a velocity-time graph.

Correct Answer: B

Solution:

For constant acceleration, the position-time graph is a parabola.

A straight line parallel to the time axis on a velocity-time graph.

A parabola opening upwards on a position-time graph.

A straight line with a positive slope on a position-time graph.

A horizontal line on an acceleration-time graph.

Correct Answer: A

Solution:

A straight line parallel to the time axis on a velocity-time graph indicates constant velocity, as there is no change in velocity over time.

44.1 m

66.15 m

88.2 m

98.0 m

Correct Answer: B

Solution:

To find the maximum height, use the equation

v^2 = u^2 + 2a x

. At the maximum height, the final velocity

v = 0

. The initial velocity

u = 29.4 \text{ m/s}

and acceleration

a = -9.8 \text{ m/s}^2

. Solving for

x

, we get

x = \frac{(0)^2 - (29.4)^2}{2 \times (-9.8)} = 44.1 \text{ m}

75 \text{ m}

50 \text{ m}

100 \text{ m}

125 \text{ m}

Correct Answer: A

Solution:

Using the equation

x = v_0 t + \frac{1}{2} a t^2

, where

v_0 = 10 \text{ m/s}

a = 2 \text{ m/s}^2

, and

t = 5 \text{ s}

, we find

x = 10 \times 5 + \frac{1}{2} \times 2 \times 5^2 = 50 + 25 = 75 \text{ m}

10 m/s²

8 m/s²

12 m/s²

14 m/s²

Correct Answer: A

Solution:

Acceleration is the derivative of velocity with respect to time.

a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t + 1) = 6t - 2

. At

t = 2

seconds,

a(2) = 6(2) - 2 = 10 \text{ m/s}^2

30 m/s

40 m/s

50 m/s

60 m/s

Correct Answer: B

Solution:

Using the equation of motion

v = v_0 + at

, where

v_0 = 20 \text{ m/s}

a = 2 \text{ m/s}^2

, and

t = 5 \text{ s}

, we find the final velocity

v = 20 + (2 \times 5) = 30 \text{ m/s}

A particle with zero speed at an instant must have zero acceleration at that instant.

A particle with zero speed may have non-zero velocity.

A particle with constant speed must have zero acceleration.

A particle with positive acceleration must be speeding up.

Correct Answer: C

Solution:

A particle with constant speed has no change in velocity, thus its acceleration is zero.

1 \text{ m/s}^2

2 \text{ m/s}^2

3 \text{ m/s}^2

4 \text{ m/s}^2

Correct Answer: A

Solution:

Using the equation

v^2 = v_0^2 + 2ax

, where

v = 10 \text{ m/s}

v_0 = 0

, and

x = 50 \text{ m}

, we find

100 = 0 + 2a \times 50

. Solving for

a

, we get

a = 1 \text{ m/s}^2

48 km/h

50 km/h

52 km/h

54 km/h

Correct Answer: A

Solution:

The average speed is calculated as the total distance divided by the total time. Let the distance between A and B be

d

. The time for the first leg is

\frac{d}{60}

and for the return leg is

\frac{d}{40}

. The total distance is

2d

and the total time is

\frac{d}{60} + \frac{d}{40} = \frac{5d}{120}

. The average speed is

\frac{2d}{\frac{5d}{120}} = 48

km/h.

1 s

2 s

3 s

4 s

Correct Answer: B

Solution:

At the highest point, the final velocity

v = 0

. Using

v = v_0 - gt

, where

v_0 = 20 \text{ m/s}

and

g = 9.8 \text{ m/s}^2

0 = 20 - 9.8t \Rightarrow t = \frac{20}{9.8} \approx 2 \text{ s}

Acceleration is zero.

Acceleration is positive.

Acceleration is negative.

Acceleration is constant but non-zero.

Correct Answer: A

Solution:

If a particle is moving with constant speed, its acceleration is zero because there is no change in velocity.

Acceleration must be zero.

Acceleration must be non-zero.

Acceleration can be zero or non-zero.

Acceleration must be negative.

Correct Answer: C

Solution:

A particle can have zero speed at an instant and still have non-zero acceleration, as in the case of a particle at the top of its trajectory when thrown upwards.

0 m/s

9.8 m/s

29.4 m/s

19.6 m/s

Correct Answer: A

Solution:

At the highest point of its motion, the velocity of the ball is 0 m/s because it momentarily stops before descending.

6 km/h

7.5 km/h

8 km/h

10 km/h

Correct Answer: A

Solution:

Total distance is

2 \times 15 = 30

km. Time taken for the first half is

\frac{15}{10} = 1.5

hours, and for the return is

\frac{15}{5} = 3

hours. Total time is

4.5

hours. Average speed is

\frac{30}{4.5} = 6 \text{ km/h}

The velocity-time graph is a straight line parallel to the time axis.

The position-time graph is a parabola.

The velocity-time graph is a curve.

The acceleration-time graph is a horizontal line at zero.

Correct Answer: B

Solution:

For uniformly accelerated motion, the position-time graph is a parabola, indicating constant acceleration.

2 m/s²

3 m/s²

4 m/s²

5 m/s²

Correct Answer: A

Solution:

Using the equation

v = v_0 + at

, where initial velocity

v_0 = 0

, final velocity

v = 30 \text{ m/s}

, and time

t = 10 \text{ s}

, we solve for acceleration

a

30 = 0 + 10a \Rightarrow a = 3 \text{ m/s}^2

Approximately 37 m/s

Approximately 27 m/s

Approximately 47 m/s

Approximately 57 m/s

Correct Answer: B

Solution:

The speed of the ball just before hitting the floor for the first time can be calculated using the equation

v = \sqrt{2gh}

. After the first collision, the speed is reduced to 90% of its original value. The speed just before the second collision is 90% of the speed just before the first collision.

5 m/s

10 m/s

15 m/s

20 m/s

Correct Answer: D

Solution:

Using the equation

v = v_0 + at

, where initial velocity

v_0 = 0

a = 2 \text{ m/s}^2

, and

t = 5 \text{ s}

, we find

v = 0 + 2 \times 5 = 10 \text{ m/s}

10 km/h

15 km/h

12.5 km/h

20 km/h

Correct Answer: C

Solution:

The total distance is 5 km (2.5 km each way). The time taken is

\frac{2.5}{5} + \frac{2.5}{25} = 0.5 + 0.1 = 0.6

hours. Average speed = total distance / total time =

\frac{5}{0.6} \approx 8.33

km/h.

It must have zero acceleration.

It may have non-zero acceleration.

It must be at rest permanently.

It must be moving backwards.

Correct Answer: B

Solution:

A particle with zero speed at an instant can have non-zero acceleration, such as at the topmost point of a projectile's path.

Displacement

Acceleration

Speed

Time

Correct Answer: A

Solution:

The area under a velocity-time graph represents the displacement of the object over the given time interval.

30 \text{ s}

45 \text{ s}

60 \text{ s}

15 \text{ s}

Correct Answer: A

Solution:

Let the time taken be

t

. The distance covered by Car A is

30t

. For Car B,

x = \frac{1}{2} \times 2 \times t^2 = t^2

. Setting

30t = t^2

, we get

t^2 - 30t = 0

, which gives

t(t - 30) = 0

. Thus,

t = 30 \text{ s}

A straight line inclined to the time axis in a velocity-time graph.

A parabolic curve in a position-time graph.

A horizontal line in a velocity-time graph.

A sinusoidal curve in a position-time graph.

Correct Answer: A

Solution:

A straight line inclined to the time axis in a velocity-time graph indicates constant acceleration, as the slope of the line represents acceleration.

Upwards

Downwards

Horizontal

None of the above

Correct Answer: B

Solution:

During the upward motion of the ball, the direction of acceleration due to gravity is downwards.

1 m/s²

2 m/s²

3 m/s²

4 m/s²

Correct Answer: B

Solution:

Acceleration is calculated using

a = \frac{v - v_0}{t}

. Here,

v = 15 \text{ m/s}

v_0 = 5 \text{ m/s}

, and

t = 5 \text{ s}

. Thus,

a = \frac{15 - 5}{5} = 2 \text{ m/s}^2

2.04 s

1.02 s

4.08 s

3.06 s

Correct Answer: A

Solution:

At maximum height, the final velocity

v = 0

. Using the equation:

v = v_0 - gt

, where

v_0 = 20 \text{ m/s}

and

g = 9.8 \text{ m/s}^2

, we solve for

t

0 = 20 - 9.8t \Rightarrow t = \frac{20}{9.8} \approx 2.04 \text{ s}

5.5 km/h

6 km/h

6.25 km/h

6.5 km/h

Correct Answer: B

Solution:

The total distance is 5 km (2.5 km each way). The total time is

\frac{2.5}{5} + \frac{2.5}{7.5} = 0.5 + \frac{1}{3} = \frac{5}{6}

hours. Average speed = Total distance / Total time =

\frac{5}{\frac{5}{6}} = 6 \text{ km/h}

20 m

30 m

40 m

50 m

Correct Answer: B

Solution:

The time to fall 45 m is found using

h = \frac{1}{2}gt^2

, giving

t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{9.8}} \approx 3 \text{ s}

. The horizontal distance is

d = vt = 10 \times 3 = 30 \text{ m}

5 km

6 km

7 km

8 km

Correct Answer: A

Solution:

The displacement can be found using the Pythagorean theorem:

\text{Displacement} = \sqrt{3^2 + 4^2} = 5 \text{ km}

12 m/s

16 m/s

22 m/s

26 m/s

Correct Answer: D

Solution:

Using the equation

v = u + at

, where

u = 10

m/s,

a = 2

m/s², and

t = 6

s, we get

v = 10 + 2 \times 6 = 22

m/s.

2 \text{ m/s}^2

4 \text{ m/s}^2

3 \text{ m/s}^2

5 \text{ m/s}^2

Correct Answer: B

Solution:

Using the equation

v^2 = v_0^2 + 2ax

, where

v = 0

v_0 = 20 \text{ m/s}

, and

x = 100 \text{ m}

, we find

0 = 400 + 2a \times 100

. Solving for

a

, we get

a = -4 \text{ m/s}^2

. The magnitude of deceleration is

4 \text{ m/s}^2

4 m/s²

8 m/s²

16 m/s²

20 m/s²

Correct Answer: A

Solution:

Using the equation of motion,

s = ut + \frac{1}{2}at^2

, where initial velocity

u = 0

s = 100

m, and

t = 5

s. Solving for

a

, we get

100 = 0 + \frac{1}{2}a(5)^2

, which gives

a = 4

m/s².

A railway carriage moving without jerks between two stations.

A monkey sitting on top of a man cycling smoothly on a circular track.

A spinning cricket ball that turns sharply on hitting the ground.

A tumbling beaker that has slipped off the edge of a table.

Correct Answer: A

Solution:

A railway carriage moving without jerks can be considered as a point object because its size is much smaller than the distance it covers.

Acceleration is in the opposite direction of velocity.

Acceleration is in the same direction as velocity.

Acceleration is perpendicular to velocity.

Acceleration direction is independent of velocity.

Correct Answer: B

Solution:

When a particle is speeding up, acceleration is in the same direction as velocity.

A railway carriage moving without jerks between two stations.

A monkey sitting on top of a man cycling smoothly on a circular track.

A spinning cricket ball that turns sharply on hitting the ground.

A tumbling beaker that has slipped off the edge of a table.

Correct Answer: A

Solution:

A railway carriage moving without jerks can be considered a point object as its size is negligible compared to the distance it travels.

150 m/s

120 m/s

180 m/s

210 m/s

Correct Answer: B

Solution:

Relative speed of the bullet with respect to the van is 150 m/s. The relative speed of the thief's car with respect to the van is 192 km/h = 53.33 m/s. Thus, the speed of the bullet relative to the thief's car is 150 m/s - 53.33 m/s = 96.67 m/s.

11.5 \text{ m}

12.0 \text{ m}

11.0 \text{ m}

10.5 \text{ m}

Correct Answer: A

Solution:

Using the equation

v^2 = v_0^2 + 2a x

, where

v = 0

v_0 = 15 \text{ m/s}

a = -9.8 \text{ m/s}^2

, we find

0 = 225 - 19.6x

. Solving for

x

, we get

x = \frac{225}{19.6} = 11.5 \text{ m}

0 km/h

1 km/h

2 km/h

3 km/h

Correct Answer: A

Solution:

The total displacement is zero as he returns home, hence the magnitude of average velocity is 0 km/h.

2 m/s

3 m/s

4 m/s

5 m/s

Correct Answer: C

Solution:

Average speed is calculated as total path length divided by time interval. Therefore, average speed = 100 m / 20 s = 5 m/s.

3.5 m/s²

4.5 m/s²

5.5 m/s²

6.5 m/s²

Correct Answer: A

Solution:

Using the equation

v^2 = u^2 + 2as

, where final velocity

v = 0

, initial velocity

u = 126 \times \frac{5}{18} \text{ m/s}

, and

s = 200 \text{ m}

, we find

a = -3.5 \text{ m/s}^2

10 m/s

15 m/s

20 m/s

25 m/s

Correct Answer: B

Solution:

Average speed is calculated as total distance divided by total time. Therefore, the average speed is

\frac{150 \text{ m}}{10 \text{ s}} = 15 \text{ m/s}

10 km/h

15 km/h

5 km/h

7.5 km/h

Correct Answer: C

Solution:

The total distance covered is 5 km (2.5 km each way) and the total time taken is 8 hours. Average speed is total distance divided by total time, which is 5 km/h.

60 m/s

80 m/s

100 m/s

120 m/s

Correct Answer: A

Solution:

The relative speed of the bullet with respect to the thief's car is given by the difference in their speeds. Convert the speeds to m/s: police van speed = 30 \times \frac{1000}{3600} \text{ m/s}, thief's car speed = 192 \times \frac{1000}{3600} \text{ m/s}. The speed of the bullet relative to the thief's car is

150 - (192 - 30) \times \frac{1000}{3600} = 60 \text{ m/s}

50 m

100 m

150 m

200 m

Correct Answer: A

Solution:

Using the kinematic equation for uniformly accelerated motion,

x = v_0 t + \frac{1}{2} a t^2

where

v_0 = 0

v = 20 \text{ m/s}

, and

t = 5 \text{ s}

. First, find acceleration

a

using

v = v_0 + at \Rightarrow a = \frac{v}{t} = \frac{20}{5} = 4 \text{ m/s}^2

. Then, calculate displacement:

x = 0 + \frac{1}{2} \times 4 \times 5^2 = 50 \text{ m}

Instantaneous speed is always greater than magnitude of velocity.

Instantaneous speed is always less than magnitude of velocity.

Instantaneous speed is always equal to magnitude of velocity.

Instantaneous speed can be greater or less than magnitude of velocity.

Correct Answer: C

Solution:

Instantaneous speed is always equal to the magnitude of instantaneous velocity, as speed is the absolute value of velocity.

6 km/h

5.5 km/h

6.25 km/h

5 km/h

Correct Answer: A

Solution:

Total distance = 5 km. Time to market = 2.5/5 = 0.5 hours. Time to return = 2.5/7.5 = 0.33 hours. Total time = 0.5 + 0.33 = 0.83 hours. Average speed = Total distance / Total time = 5 / 0.83 ≈ 6 km/h.

15 km/h

20 km/h

25 km/h

30 km/h

Correct Answer: B

Solution:

Average speed is calculated as total distance divided by total time. Here,

\text{Average speed} = \frac{10 \text{ km}}{0.5 \text{ hours}} = 20 \text{ km/h}

3.5 m/s²

4.5 m/s²

5.5 m/s²

6.5 m/s²

Correct Answer: A

Solution:

To find the retardation, we use the equation

v^2 = u^2 + 2a x

. Here, the final velocity

v = 0

, initial velocity

u = 126 \times \frac{1000}{3600} \text{ m/s}

, and

x = 200 \text{ m}

. Solving for

a

, we get

a = -\frac{(126 \times \frac{1000}{3600})^2}{2 \times 200} = -3.5 \text{ m/s}^2

. The magnitude of retardation is 3.5 m/s².

4 seconds

5 seconds

6 seconds

7 seconds

Correct Answer: A

Solution:

Using the equation

v = v_0 + at

, where final velocity

v = 0

, initial velocity

v_0 = 20 \text{ m/s}

, and acceleration

a = -5 \text{ m/s}^2

, we solve for time

t

0 = 20 - 5t \Rightarrow t = 4 \text{ seconds}

27 m/s

18 m/s

33 m/s

12 m/s

Correct Answer: A

Solution:

To find the final velocity, use the equation:

v = v_0 + at

where

v_0 = 15 \text{ m/s}

a = 3 \text{ m/s}^2

, and

t = 4 \text{ s}

. Thus,

v = 15 + 3 \times 4 = 27 \text{ m/s}

2 m/s²

4 m/s²

6 m/s²

8 m/s²

Correct Answer: B

Solution:

Centripetal acceleration

a_c

is given by

a_c = \frac{v^2}{r}

, where

v = 20 \text{ m/s}

and

r = 100 \text{ m}

. Thus,

a_c = \frac{20^2}{100} = 4 \text{ m/s}^2

11.5 m

12.5 m

13.5 m

14.5 m

Correct Answer: B

Solution:

Using the equation

v^2 = v_0^2 + 2a x

, where final velocity

v = 0

, initial velocity

v_0 = 15 \text{ m/s}

, and acceleration

a = -9.8 \text{ m/s}^2

, we solve for height

x

0 = 15^2 - 2 \times 9.8 \times x \Rightarrow x = 11.5 \text{ m}

32 m

40 m

48 m

56 m

Correct Answer: D

Solution:

Using the equation

x = v_0t + \frac{1}{2}at^2

, where

v_0 = 3 \text{ m/s}

a = 4 \text{ m/s}^2

, and

t = 4 \text{ s}

, we find

x = 3 \times 4 + \frac{1}{2} \times 4 \times 16 = 56 \text{ m}

If the particle is speeding up, acceleration is in the direction of velocity.

If the particle is speeding up, acceleration is opposite to the direction of velocity.

Acceleration is always zero when the particle is speeding up.

Acceleration is always negative when the particle is speeding up.

Correct Answer: A

Solution:

Solution:

Instantaneous speed is the absolute value of instantaneous velocity, so they are equal in magnitude.

Motion in a Straight Line

AI Learning Assistant

Summary

Summary of Motion in a Straight Line

Learning Objectives

Learning Objectives

Detailed Notes

Chapter 2: Motion in a Straight Line

2.1 Introduction

2.2 Instantaneous Velocity and Speed

2.3 Acceleration

2.4 Kinematic Equations for Uniformly Accelerated Motion

2.5 Relative Velocity

Points to Ponder

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Exam Preparation

Practice & Assessment

Multiple Choice Questions

Q1.In which of the following examples of motion, can the body be considered approximately a point object?

Correct Answer: A

Q2.A ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height reached by the ball? (Take g=9.8g = 9.8g=9.8 m/s²)

Correct Answer: B

Q3.A police car is chasing a thief's car. The police car is moving at 30 km/h and fires a bullet at a speed of 150 m/s. If the thief's car is moving at 192 km/h, what is the speed of the bullet relative to the thief's car?

Correct Answer: A

Q4.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 second. How long does the drunkard take to fall in a pit 13 m away from the start?

Correct Answer: B

Q5.A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, each step being 1 m long and taking 1 second. How long does it take for the drunkard to fall in a pit 13 m away from the start?

Correct Answer: C

Q6.In which of the following examples of motion, can the body be considered approximately a point object?

Correct Answer: A

Q7.A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. What is the speed of the ball just before it hits the floor for the second time?

Correct Answer: B

Q8.A particle moves along a straight line with a velocity given by v(t)=4t−3v(t) = 4t - 3v(t)=4t−3. What is the displacement of the particle from t=0t = 0t=0 to t=5t = 5t=5 seconds?

Correct Answer: A

Q9.A ball is thrown upwards with an initial speed of 29.4 m/s. What is the direction of acceleration during the upward motion of the ball?

Correct Answer: B

Q10.A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. What is the average speed of the man over the entire journey?

Correct Answer: A

Q11.A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. What is the speed of the ball just before it hits the floor for the first time? (Assume g=9.8 m/s2g = 9.8 \text{ m/s}^2g=9.8 m/s2 and no air resistance.)

Correct Answer: B

Q12.Which graph correctly represents the motion of a particle with constant acceleration?

Correct Answer: B

Q13.Which of the following graphs represents a particle moving with constant velocity?

Correct Answer: A

Q14.A player throws a ball upwards with an initial speed of 29.4 m/s. Assuming no air resistance, what is the maximum height reached by the ball?

Correct Answer: B

Q15.A particle moves along a straight line with an initial velocity of 10 m/s10 \text{ m/s}10 m/s and a constant acceleration of 2 m/s22 \text{ m/s}^22 m/s2. How far does the particle travel in the first 5 seconds5 \text{ seconds}5 seconds?

Correct Answer: A

Q16.A particle moves along a straight line with a velocity given by v(t)=3t2−2t+1v(t) = 3t^2 - 2t + 1v(t)=3t2−2t+1. What is the acceleration of the particle at t=2t = 2t=2 seconds?

Correct Answer: A

Q17.A car is moving along a straight road with an initial velocity of 20 m/s. It accelerates uniformly at 2 m/s² for 5 seconds. What is the final velocity of the car?

Correct Answer: B

Q18.A particle is moving in one-dimensional motion. Which of the following statements is true?

Correct Answer: C

Q19.A cyclist accelerates uniformly from rest to a speed of 10 m/s10 \text{ m/s}10 m/s over a distance of 50 m50 \text{ m}50 m. What is the acceleration of the cyclist?

Correct Answer: A

Q20.A train travels from station A to station B with a speed of 60 km/h and returns to station A with a speed of 40 km/h. What is the average speed of the train for the entire journey?

Correct Answer: A

Q21.A ball is thrown vertically upwards with a speed of 20 m/s. How long does it take to reach the highest point of its trajectory?

Correct Answer: B

Q22.If a particle is moving with constant speed, what can be said about its acceleration?

Correct Answer: A

Q23.In one-dimensional motion, if a particle has zero speed at an instant, what can be said about its acceleration at that instant?

Correct Answer: C

Q24.A player throws a ball upwards with an initial speed of 29.4 m/s. What is the velocity of the ball at the highest point of its motion?

Correct Answer: A

Q25.A cyclist travels a distance of 15 km with a speed of 10 km/h and returns with a speed of 5 km/h. What is the average speed of the cyclist for the entire trip?

Correct Answer: A

Q26.Which of the following statements is true for an object in uniformly accelerated motion?

Correct Answer: B

Q27.A train accelerates uniformly from rest and reaches a speed of 30 m/s in 10 seconds. What is the acceleration of the train?

Correct Answer: A

Q28.A ball is dropped from a height of 90 m. After each collision with the floor, the ball loses one tenth of its speed. What is the speed of the ball just before its second collision with the floor?

Correct Answer: B

Q29.A train moves from rest with a constant acceleration of 2 m/s². What will be its velocity after 5 seconds?

Correct Answer: D

Q30.What is the average speed of a woman who walks to her office 2.5 km away at 5 km/h and returns home by an auto at 25 km/h?

Correct Answer: C