Chapter 3: Motion in a Straight Line

1. Introduction to Kinematics

The study of motion is a foundational pillar of physics. The branch of physics dedicated to this study is known as Mechanics. Mechanics is systematically divided into two sub-branches:

• Kinematics: This is the study of the geometry of motion. In kinematics, we describe *how* objects move without concerning ourselves with the forces that cause the motion. We utilize quantities like position, displacement, velocity, and acceleration to construct a mathematical framework of the motion.
• Dynamics: This sub-branch investigates the cause of motion, which is force. It connects motion to the forces that produce it and is fundamentally based on Newton's laws of motion.

This chapter is an introduction to Kinematics, focusing specifically on the simplest case: motion in a single dimension, also known as motion in a straight line or rectilinear motion.

2. Frame of Reference and the Concept of Rest and Motion

An object is in motion if it changes its position with respect to its surroundings as time passes. It is at rest if its position does not change with respect to its surroundings. However, these concepts are **relative**. A passenger on a moving train is at rest relative to fellow passengers but in motion relative to the ground. This relativity necessitates a clear point of view from which to observe motion.

2.1 Point Object (Particle)

In kinematics, real objects are often idealized as point objects or particles. This simplification is valid when the distance an object travels is significantly larger than its own dimensions. For example, a car traveling from Delhi to Mumbai can be treated as a point object.

2.2 Motion in One, Two, and Three Dimensions

The motion of an object is classified based on the number of coordinates required to specify its position:

• One-dimensional motion (1-D): Only one coordinate is required. Example: a train moving on a straight track.
• Two-dimensional motion (2-D): Two coordinates are required. Example: a car on a curved road.
• Three-dimensional motion (3-D): Three coordinates are required. Example: a bird flying in the sky.

This chapter focuses on **one-dimensional motion**.

3. Path Length (Distance) and Displacement

To quantitatively describe motion along a straight line, we use two key concepts: path length and displacement. Let's consider an object moving along an x-axis, starting from an origin O.

3.1 Path Length (Distance)

3.2 Displacement

4. Uniform and Non-Uniform Motion

4.1 Uniform Motion

An object is said to be in uniform motion if it covers equal displacements in equal intervals of time, however small these intervals may be. In uniform motion, the velocity of the object is constant, both in magnitude and direction. This implies that the acceleration of the object is zero.

4.2 Non-Uniform Motion

An object is in non-uniform motion if it covers unequal displacements in equal intervals of time. In this case, the velocity of the object is variable (it changes with time). Most motions in real life are non-uniform.

Example: A car driving in city traffic repeatedly stops, speeds up, and slows down. It covers unequal distances in every second. This is non-uniform motion. Conversely, a car driving on a straight empty highway at a steady 80 km/h with cruise control engaged is an example of uniform motion.

5. Average Speed and Average Velocity

For non-uniform motion, we use average values to describe the overall rate of motion.

5.1 Average Speed

5.2 Average Velocity

6. Elementary Concepts of Calculus for Describing Motion

While average velocity is useful, it doesn't describe how fast an object is moving at a specific moment. The reading on a car's speedometer provides its instantaneous speed. To define this concept mathematically, we need the powerful tool of **calculus**.

6.1 Instantaneous Velocity (The Concept of Differentiation)

To find the velocity "at an instant," we must consider the average velocity over an infinitesimally small time interval (Δt → 0). This limiting value of the average velocity is the instantaneous velocity.

7. Acceleration

In non-uniform motion, velocity changes with time. The rate at which velocity changes is called acceleration.

7.1 Average and Instantaneous Acceleration

Nature of Acceleration:

• Acceleration is a vector quantity.
• Positive Acceleration: Velocity is increasing in the positive direction.
• Negative Acceleration (Retardation): Velocity is decreasing in the positive direction.
• SI Unit: The SI unit of acceleration is metre per second squared (m/s²).

7.2 Elementary Concepts of Integration

Integration is the inverse of differentiation. In kinematics, it allows us to find displacement from a velocity function and velocity from an acceleration function. Geometrically, integration represents the **area under a curve**.

8. Graphical Representation of Motion

Graphs provide a powerful visual tool for analyzing motion. The shape, slope, and area of these graphs carry significant physical meaning.

8.1 Position-Time (x-t) Graph

This graph plots the position (x) of an object against time (t).

8.2 Velocity-Time (v-t) Graph

This graph plots the velocity (v) of an object against time (t).

9. Kinematic Equations for Uniformly Accelerated Motion (Graphical Treatment)

For an object moving along a straight line with uniform acceleration 'a', we can derive three fundamental equations relating initial velocity (u), final velocity (v), time (t), and displacement (s).

Consider an object starting with initial velocity 'u' at t=0 and reaching a final velocity 'v' at time 't'. Its v-t graph will be a straight line as shown below.

9.1 First Equation of Motion: Velocity-Time Relation

The acceleration 'a' is the slope of the v-t graph.

Rearranging gives the first kinematic equation:

9.2 Second Equation of Motion: Position-Time Relation

The displacement 's' is the area under the v-t graph (area of trapezoid OABD).

This gives the second kinematic equation:

9.3 Third Equation of Motion: Position-Velocity Relation

This equation is derived by eliminating time 't' from the first two equations, or directly from the area of the trapezoid.

Rearranging gives the third kinematic equation:

10. Kinematic Equations for Uniformly Accelerated Motion (Calculus Method)

Calculus provides a more elegant and powerful method to derive the equations of motion. We assume that the object moves with a constant acceleration a.

10.1 First Equation: Velocity-Time Relation

By definition, instantaneous acceleration is the rate of change of velocity:

Integrating both sides with proper limits: at t = 0, velocity is u; at time t = t, velocity is v.

Since acceleration a is constant, we can take it outside the integral:

10.2 Second Equation: Position-Time Relation

By definition, instantaneous velocity is the rate of change of position:

Substitute v = u + at from the first equation:

Integrating both sides with limits: at t = 0, displacement s = 0; at time t = t, displacement is s.

10.3 Third Equation: Velocity-Position Relation

We start with the definition of acceleration: a = dv / dt.

Using the chain rule, we can write: a = (dv / ds) × (ds / dt)

Since ds / dt = v, this becomes:

Integrating both sides with limits: when displacement is 0, velocity is u; when displacement is s, velocity is v.

Since a is constant:

Important Questions and Answers

Conceptual Questions

Q1: What is a frame of reference and why is it essential for describing motion?

A: A frame of reference is a coordinate system (like x, y, z axes) combined with a clock, with respect to which an observer describes the motion of an object. It is essential because motion and rest are relative concepts; an object's state of motion can only be defined relative to an observer or a specific reference frame.

Q2: Distinguish between distance and displacement with a suitable example.

A: Distance is the total path length covered by an object (a scalar quantity), while displacement is the shortest straight-line distance between the initial and final positions (a vector quantity).
Example: If a person walks 4 m east and then 3 m north, the distance covered is 4 m + 3 m = 7 m. The displacement is the hypotenuse of the right triangle formed, which is √(4² + 3²) = √25 = 5 m, in the northeast direction.

Q3: Can an object have a constant speed but a varying velocity? Give an example.

A: Yes. Velocity is a vector quantity, possessing both magnitude (speed) and direction. An object can have a constant speed while its direction of motion is changing. Since velocity changes if either speed or direction changes, its velocity will be varying.
Example: An object in uniform circular motion, like a satellite orbiting the Earth at a constant speed, has a continuously changing velocity because its direction of motion is changing at every instant.

Q4: The displacement of an object is zero, but the distance covered by it is not zero. Is this situation possible? Explain with an example.

A: Yes, this situation is possible.
Example: An athlete running one complete lap on a 400 m circular track. The total distance covered is 400 m. However, since the athlete returns to the starting point, the initial and final positions are the same. Therefore, the displacement is zero.

Q5: What do the slope of a position-time graph and the slope of a velocity-time graph represent physically?

A: The slope of a position-time (x-t) graph represents the instantaneous velocity of the object. The slope of a velocity-time (v-t) graph represents the instantaneous acceleration of the object.

Q6: What physical quantity is represented by the area under a velocity-time graph?

A: The area under a velocity-time (v-t) graph represents the displacement of the object over that time interval.

Q7: Can a body have zero velocity and still be accelerating? Give an example.

A: Yes. An object can have a momentary zero velocity while still being under constant acceleration.
Example: When a ball is thrown vertically upwards, its velocity becomes zero at the highest point of its trajectory. However, the acceleration due to gravity (g ≈ 9.8 m/s²) is still acting on it, causing it to fall back down. Thus, at the highest point, velocity is zero but acceleration is non-zero.

Q8: The kinematic equations of motion are valid only under what specific condition?

A: The three standard kinematic equations of motion (v = u + at, etc.) are valid only for motion in a straight line with uniform (constant) acceleration.

Q9: Why are SI units considered a coherent system of units?

A: The SI system is considered coherent because all derived units can be obtained by simple multiplication or division of the base units without the need for any numerical conversion factors. For example, the unit of force (Newton) is directly derived as kg × m / s², with a proportionality constant of 1.

Q10: Draw position-time graphs for two objects having zero relative velocity.

A: Two objects have zero relative velocity if they are moving with the same constant velocity. On a position-time graph, this would be represented by two parallel straight lines. The slope of both lines would be the same, indicating equal velocities, but their intercepts on the position axis could be different, indicating they started from different positions.

Numerical Problems

Q11: The position of an object moving along x-axis is given by x = a + bt², where a = 8.5 m and b = 2.5 m s⁻². What is its velocity at t = 2.0 s?

A:
Given: x(t) = 8.5 + 2.5t².
Velocity is the first derivative of position with respect to time: v(t) = dx/dt.
v(t) = d/dt (8.5 + 2.5t²) = 0 + (2.5 × 2t) = 5.0t.
At t = 2.0 s: v(2.0) = 5.0 × 2.0 = 10 m/s.
Answer: The velocity at t = 2.0 s is 10 m/s.

Q12: A ball is thrown vertically upwards with a velocity of 20 m/s from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise? and (b) how long will it be before the ball hits the ground? Take g = 10 m/s².

A:
Given: u = +20 m/s, g = -10 m/s², initial height = 25 m.
(a) How high will the ball rise?
At the highest point, v = 0. Using v² = u² + 2as:
0² = 20² + 2(-10)s → 0 = 400 - 20s → 20s = 400 → s = 20 m.
The ball rises 20 m above the point of throw.

(b) How long before it hits the ground?
We can use s = ut + (1/2)at² for the entire journey from the top of the building to the ground.
Here, displacement s = -25 m (since it ends up 25 m below the starting point).
-25 = 20t + (1/2)(-10)t² → -25 = 20t - 5t²
5t² - 20t - 25 = 0 → t² - 4t - 5 = 0
Factoring: (t - 5)(t + 1) = 0.
Since time cannot be negative, t = 5 s.
Answer: (a) The ball will rise 20 m. (b) It will hit the ground after 5 s.

Q13: A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, find the maximum velocity reached.

A:
Let t₁ be the time of acceleration and t₂ be the time of deceleration.
Total time t = t₁ + t₂.
Let v_max be the maximum velocity.
For acceleration phase: vmax = 0 + αt1 → t1 = vmax / α.
For deceleration phase: 0 = vmax - βt2 → t2 = vmax / β.
Substitute t₁ and t₂ into the total time equation:
t = (vmax / α) + (vmax / β) = vmax(1/α + 1/β) = vmax((α+β)/αβ).
Solving for v_max: vmax = (αβ / (α+β))t.
Answer: The maximum velocity reached is (αβ / (α+β))t.

Q14: 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. Determine how long the drunkard takes to fall in a pit 13 m away from the start.

A:
In each cycle of 5 steps forward + 3 steps backward:
Time taken = 5 s + 3 s = 8 s.
Net displacement = 5 m - 3 m = 2 m.
So, the drunkard moves a net of 2 m every 8 s.
To reach 8 m, he will take 4 cycles: 4 × 8 s = 32 s. At this point, he is at x = 8 m.
In the next cycle, he takes 5 steps forward.
Displacement = 8 m + 5 m = 13 m.
Time taken for these 5 steps = 5 s.
Total time = Time to reach 8 m + Time for last 5 steps = 32 s + 5 s = 37 s.
At 37 s, he reaches the pit at 13 m and falls in. He does not take the 3 steps backward.
Answer: The drunkard takes 37 s to fall in the pit.

Q15: A body travels a distance s₁ with velocity v₁ and distance s₂ with velocity v₂. Calculate the average speed.

A:
Time for first part (t₁) = s1 / v1.
Time for second part (t₂) = s2 / v2.
Total Distance = s1 + s2.
Total Time = t1 + t2 = (s1/v1) + (s2/v2).
Average Speed = Total Distance / Total Time = (s1 + s2) / ((s1/v1) + (s2/v2)).
Simplifying the denominator: (s1v2 + s2v1) / v1v2.
Average Speed = (s1 + s2)v1v2 / (s1v2 + s2v1).
Answer: Average Speed = (s₁ + s₂)v₁v₂ / (s₁v₂ + s₂v₁).

Q16: A car moving with a speed of 50 km/h can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/h, what is the minimum stopping distance?

A:
Let the braking force (and thus retardation 'a') be constant.
Using v² = u² + 2as, the stopping distance s = -u² / 2a.
This shows that stopping distance is proportional to the square of the initial speed (s ∝ u²).
Let u₁ = 50 km/h, s₁ = 6 m.
Let u₂ = 100 km/h, s₂ = ?
Since u₂ = 2u₁, then s₂ must be (2)² = 4 times s₁.
s2 = 4 × s1 = 4 × 6 m = 24 m.
Answer: The minimum stopping distance is 24 m.

Q17: The velocity of a particle is given by v(t) = 3t² - 2t + 5 m/s. Find the displacement of the particle between t = 1 s and t = 3 s.

A:
Displacement is the integral of velocity with respect to time.
Displacement (Δx) = ∫13 v(t) dt = ∫13 (3t² - 2t + 5) dt
= [3(t³/3) - 2(t²/2) + 5t]13
= [t³ - t² + 5t]13
= [(3)³ - (3)² + 5(3)] - [(1)³ - (1)² + 5(1)]
= [27 - 9 + 15] - [1 - 1 + 5]
= [33] - [5] = 28 m.
Answer: The displacement is 28 m.

Q18: A stone is dropped from a cliff. Find the ratio of the distances travelled by it in the first, second, and third second of its motion.

A: This is a case of free fall, u=0. We use the formula for distance in the nth second: sn = u + g/2 (2n - 1). Since u=0, sn ∝ (2n - 1).
Distance in 1st second (s₁) = g/2 (2(1) - 1) = g/2.
Distance in 2nd second (s₂) = g/2 (2(2) - 1) = 3g/2.
Distance in 3rd second (s₃) = g/2 (2(3) - 1) = 5g/2.
The ratio s₁ : s₂ : s₃ = (g/2) : (3g/2) : (5g/2) = 1 : 3 : 5.
Answer: The ratio of distances is 1 : 3 : 5 (Galileo's law of odd numbers).

Q19: A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief's car speeding away in the same direction with a speed of 192 km/h. If the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief's car?

A:
Speeds relative to ground:
V_P (Police) = 30 km/h = 30 × (5/18) = 25/3 m/s.
V_T (Thief) = 192 km/h = 192 × (5/18) = 160/3 m/s.
Muzzle speed of bullet (speed relative to police van), V_BP = 150 m/s.

Speed of bullet relative to ground (V_B) = V_BP + V_P = 150 + 25/3 = (450+25)/3 = 475/3 m/s.
Speed of bullet relative to thief's car (V_BT) = V_B - V_T
VBT = (475/3) - (160/3) = 315/3 = 105 m/s.
Answer: The bullet hits the thief's car at a speed of 105 m/s.

Q20: A body covers 12 m in the 2nd second and 20 m in the 4th second. How much distance will it cover in 4 seconds after the 5th second?

A:
Using sn = u + a/2 (2n - 1):
For 2nd second: 12 = u + a/2 (2(2)-1) → 12 = u + 1.5a ---(1)
For 4th second: 20 = u + a/2 (2(4)-1) → 20 = u + 3.5a ---(2)
Subtracting (1) from (2): 8 = 2a → a = 4 m/s².
Substitute 'a' in (1): 12 = u + 1.5(4) → 12 = u + 6 → u = 6 m/s.

We need distance covered in 4 seconds after the 5th second, i.e., between t=5s and t=9s.
Distance in 9s (S₉) = ut + (1/2)at² = 6(9) + (1/2)(4)(9)² = 54 + 2(81) = 54 + 162 = 216 m.
Distance in 5s (S₅) = 6(5) + (1/2)(4)(5)² = 30 + 2(25) = 30 + 50 = 80 m.
Distance covered between 5s and 9s = S₉ - S₅ = 216 - 80 = 136 m.
Answer: It will cover 136 m.

NCERT Textbook Exercises (Detailed Solutions)

NCERT 3.1: A woman starts from her home at 9.00 am, walks with a speed of 5 km/h on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km/h. Choose suitable scales and plot the x-t graph of her motion.

Solution:
1. Journey to Office:
Distance = 2.5 km, Speed = 5 km/h.
Time taken = Distance / Speed = 2.5 / 5 = 0.5 h = 30 minutes.
She reaches the office at 9:30 am.
2. Stay at Office:
She is at rest at x = 2.5 km from 9:30 am to 5:00 pm.
3. Return Journey:
Distance = 2.5 km, Speed = 25 km/h.
Time taken = Distance / Speed = 2.5 / 25 = 0.1 h = 6 minutes.
She reaches home at 5:06 pm.
(The x-t graph would show a straight line from (9:00, 0) to (9:30, 2.5), a horizontal line to (17:00, 2.5), and a steep straight line down to (17:06, 0).)

NCERT 3.2: A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Solution:
Let the direction of the jet airplane be positive.
Velocity of jet relative to ground, vJ = +500 km/h.
Velocity of combustion products relative to jet, vPJ = -1500 km/h (since ejected backwards).
We know, vPJ = vP - vJ.
Therefore, Velocity of products relative to ground, vP = vPJ + vJ.
vP = -1500 + 500 = -1000 km/h.
Answer: The speed of the combustion products with respect to the ground is 1000 km/h (in the direction opposite to the jet).

NCERT 3.3: Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s². If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Solution:
Initial speed of both trains, u = 72 km/h = 72 × (5/18) = 20 m/s.
Let the original distance between the driver of B and the guard of A be 'x'.
Relative initial velocity of B with respect to A, uBA = uB - uA = 20 - 20 = 0.
Relative acceleration of B with respect to A, aBA = aB - aA = 1 - 0 = 1 m/s².
Time taken to overtake, t = 50 s.
When the guard of B brushes past the driver of A, train B has covered the initial distance 'x' plus the length of both trains (400m + 400m = 800m).
So, relative displacement s = x + 800.
Using relative kinematic equation: s = uBAt + (1/2)aBAt²
x + 800 = 0(50) + (1/2)(1)(50)²
x + 800 = (1/2)(2500) = 1250
x = 1250 - 800 = 450 m.
Answer: The original distance between the driver of B and guard of A was 450 m.

NCERT Textbook Exercises (Continued)

NCERT 3.4: On a two-lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Solution:
Velocity of car A, vA = 36 km/h = 10 m/s.
Velocity of car B, vB = 54 km/h = 15 m/s (moving in same direction as A).
Velocity of car C, vC = -54 km/h = -15 m/s (moving opposite to A).
Relative velocity of B wrt A, vBA = vB - vA = 15 - 10 = 5 m/s.
Relative velocity of C wrt A, vCA = |vC| + vA = 15 + 10 = 25 m/s (approaching speed).
At t = 0, distance AB = AC = 1 km = 1000 m.
Time taken by C to reach A: t = Distance / vCA = 1000 / 25 = 40 s.
Car B must overtake A within this 40 s to avoid an accident.
For car B to cover 1000 m relative to A in 40 s:
s = vBAt + (1/2)at²
1000 = 5(40) + (1/2)a(40)²
1000 = 200 + 800a
800 = 800a ⇒ a = 1 m/s².
Answer: Minimum acceleration required by car B is 1 m/s².

NCERT 3.5: Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km/h in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed do the buses ply on the road?

Solution:
Let 'v' be the speed of the bus and 'T' be the period.
Distance between two consecutive buses on the road = v × T.
Speed of cyclist, vc = 20 km/h.
1. Buses moving in same direction (A to B):
Relative velocity of bus wrt cyclist = v - vc = v - 20.
Time interval between buses passing him = 18 min = 18/60 h = 3/10 h.
Distance = Relative Speed × Time ⇒ vT = (v - 20) × (3/10) --- (Eq 1)
2. Buses moving in opposite direction (B to A):
Relative velocity of bus wrt cyclist = v + vc = v + 20.
Time interval between buses passing him = 6 min = 6/60 h = 1/10 h.
Distance = Relative Speed × Time ⇒ vT = (v + 20) × (1/10) --- (Eq 2)
Equating Eq 1 and Eq 2:
(v - 20) × (3/10) = (v + 20) × (1/10)
3(v - 20) = (v + 20)
3v - 60 = v + 20
2v = 80 ⇒ v = 40 km/h.
Substitute v in Eq 2:
40 × T = (40 + 20) × (1/10) = 60/10 = 6
T = 6/40 h = 6/40 × 60 min = 9 minutes.
Answer: Speed of buses is 40 km/h and the period T is 9 minutes.