"Motion in a Straight Line"

About

This chapter introduces kinematics — the study of motion without considering its causes. You will learn to describe the motion of objects along a straight line using physical quantities like displacement, velocity, and acceleration. The graphical analysis of motion and the concept of relative velocity are also covered.

Key Concepts

2.1 Rest and Motion

• A body is at rest if its position does not change with time relative to a reference point.
• A body is in motion if its position changes with time relative to a reference point.
• Rest and motion are relative — an object at rest relative to one observer may be in motion relative to another.

2.2 Distance and Displacement

2.3 Speed and Velocity

Speed: $\text{Average Speed}=\frac{\text{Total Distance}}{\text{Total Time}}$

Velocity: $\text{Average Velocity}=\frac{\text{Net Displacement}}{\text{Total Time}}$

$\text{Instantaneous Velocity}={\lim }_{\Delta t\to 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}$

• A body can have non-zero average speed but zero average velocity — e.g., completing a circular track and returning to the start.

2.4 Acceleration

$\text{Average Acceleration}=\frac{\Delta v}{\Delta t}$

$\text{Instantaneous Acceleration}=\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$

• Uniform acceleration: Velocity changes by equal amounts in equal time intervals.
• Zero acceleration: Velocity is constant (uniform motion).

2.5 Equations of Uniformly Accelerated Motion

For motion along a straight line with constant acceleration $a$:

1. $v=u+at$
2. $s=ut+\frac{1}{2}a{t}^2$
3. $v^2=u^2+2as$
4. $s=\left(\frac{u+v}{2}\right)t$

Where $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $t$ = time, $s$ = displacement.

2.6 Graphical Representation of Motion

Position-Time (x-t) Graph:

• Straight line with constant slope: Uniform velocity (zero acceleration)
• Curved line: Accelerated motion
• Horizontal line: Body at rest
• Slope of x-t graph = velocity

Velocity-Time (v-t) Graph:

• Slope of v-t graph = acceleration
• Area under v-t graph = displacement

2.7 Relative Velocity

The velocity of object A relative to object B: ${\vec{v}}_{AB}={\vec{v}}_A-{\vec{v}}_B$

• When two bodies move with the same velocity in the same direction, their relative velocity is zero.
• When a person moves inside a moving vehicle, the velocity relative to ground = velocity of person relative to vehicle + velocity of vehicle.

INTEXT QUESTIONS 2.1

Q1. Is it possible for a moving body to have non-zero average speed but zero average velocity during any given interval of time? If so, explain.

Ans: Yes, this is possible. Average velocity is the ratio of net displacement to total time, while average speed is the ratio of total distance traveled to total time.

Consider a body moving in a circular path and returning to its starting point. The total distance traveled is non-zero (circumference of the circle), but the net displacement is zero (as initial and final positions are the same).

Example: A car travels on a circular track of radius 100 m and completes one full circle in 60 s.

• Distance traveled = $2\pi r=2\pi (100)=628.3$ m
• Average speed = $628.3/60=10.47$ m/s
• Displacement = 0
• Average velocity = $0/60=0$ m/s

Q2. A lady drove to the market at a speed of 8 km h⁻¹. Finding market closed, she came back home at a speed of 10 km h⁻¹. If the market is 2 km away from her home, calculate the average velocity and average speed.

Ans:

• Distance to market = 2 km, speed to market = 8 km h⁻¹, speed back home = 10 km h⁻¹
• Time to reach market: $t_1=\frac{2}{8}=0.25$ h
• Time to return home: $t_2=\frac{2}{10}=0.2$ h
• Total time = $0.25+0.2=0.45$ h

$\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}=\frac{2+2}{0.45}=\frac{4}{0.45}=8.89{\text{ km h}}^{-1}$

$\text{Average velocity}=\frac{\text{Net displacement}}{\text{Total time}}=\frac{0}{0.45}=0{\text{ km h}}^{-1}$

(Since she returns to her starting point, net displacement = 0)

Q3. Can a moving body have zero relative velocity with respect to another body? Give an example.

Ans: Yes, when two bodies move with the same velocity in the same direction, their relative velocity is zero.

Example: Two cars A and B traveling on a highway, both moving at 60 km h⁻¹ in the same direction. $\text{Relative velocity of B with respect to A}=v_B-v_A=60-60=0{\text{ km h}}^{-1}$

Q4. A person strolls inside a train with a velocity of 1.0 m s⁻¹ in the direction of motion of the train. If the train is moving with a velocity of 3.0 m s⁻¹, calculate his:

(a) velocity as seen by passengers in the compartment, and

Ans: Velocity as seen by passengers in the compartment = 1.0 m s⁻¹ (This is the person's velocity relative to the train)

(b) velocity with respect to a person sitting on the platform.

Ans: Velocity with respect to a person on the platform = Velocity of person relative to train + Velocity of train $=1.0+3.0=4.0{\text{ m s}}^{-1}$

INTEXT QUESTIONS 2.2

Q1. Draw the position-time graph for a motion with zero acceleration.

Ans: For zero acceleration, velocity is constant. Therefore, position increases linearly with time. The position-time graph is a straight line with constant slope equal to the velocity.

• Positive slope → moving in positive direction
• Horizontal line → body at rest
• Negative slope → moving in negative direction

Q2. The following figure shows the displacement-time graph for two students A and B who start from their school and reach their homes. Look at the graphs carefully and answer the following questions.

(i) Do they both leave school at the same time?

Ans: Yes, both graphs start from the origin (t = 0), indicating they leave school simultaneously.

(ii) Who stays farther from the school?

Ans: Student B stays farther from school, as the final displacement of B is greater than that of A.

(iii) Do they both reach their respective houses at the same time?

Ans: Yes, both graphs end at the same time value on the x-axis.

(iv) Who moves faster?

Ans: Student A moves faster. The slope of line A (Δx/Δt) is steeper than that of line B, indicating higher average velocity.

(v) At what distance from the school do they cross each other?

Ans: They cross where the two lines intersect on the graph. Reading from the intersection point, this occurs at the displacement value where both lines meet.

Terminal Exercise

2. Define average speed and average velocity. A car travels 100 km at 50 km/h and then another 100 km at 100 km/h. Find the average speed for the whole journey.

3. Derive the equations of motion for a body moving with uniform acceleration: (a) $v=u+at$ (b) $s=ut+\frac{1}{2}a{t}^2$ (c) $v^2=u^2+2as$

4. A ball is thrown vertically upward with a velocity of 20 m/s. Calculate: (a) Maximum height reached (b) Time taken to return to the thrower's hand (Take g = 10 m/s²)

5. Draw and explain: (a) Position-time graph for uniform motion and uniformly accelerated motion (b) Velocity-time graph for uniform motion and uniformly accelerated motion

6. Prove that the area under the velocity-time graph gives the displacement.

7. Explain the concept of relative velocity. Two trains A and B of lengths 400 m each are moving on two parallel tracks with uniform speeds of 72 km/h in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between the two trains?

8. 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?

9. Explain with examples: Can an object have (a) zero velocity but non-zero acceleration? (b) constant speed but variable velocity?

10. Draw and discuss the nature of the x-t graph for a body thrown vertically upward and returning to the ground.

Worked Examples

Example 1: Average Speed

Problem: A car travels the first half of a journey at 40 km/h and the second half at 60 km/h. Find the average speed.

Solution: Let the total distance be $2d$.

• Time for first half: $t_1=\frac{d}{40}$
• Time for second half: $t_2=\frac{d}{60}$
• Total time: $t=\frac{d}{40}+\frac{d}{60}=\frac{3d+2d}{120}=\frac{5d}{120}$

$\text{Average speed}=\frac{2d}{5d/120}=\frac{2d\times 120}{5d}=48\text{ km/h}$

Example 2: Stopping Distance

Problem: A car moving at 72 km/h is brought to rest in 5 seconds by applying brakes. Find: (a) retardation, (b) distance travelled before stopping.

Solution:

• $u=72\text{ km/h}=72\times \frac{5}{18}=20\text{ m/s}$
• $v=0$, $t=5$ s

(a) $a=\frac{v-u}{t}=\frac{0-20}{5}=-4{\text{ m/s}}^2$ (Retardation = 4 m/s²)

(b) $s=ut+\frac{1}{2}a{t}^2=20(5)+\frac{1}{2}(-4)(25)=100-50=50\text{ m}$

Example 3: Free Fall

Problem: A stone is dropped from a tower 100 m high. Find (a) time to reach the ground, (b) velocity on striking the ground. (g = 10 m/s²)

Solution: $u=0$, $s=100$ m, $a=g=10$ m/s²

(a) $s=ut+\frac{1}{2}g{t}^2⟹100=0+\frac{1}{2}(10)t^2⟹t^2=20⟹t=2\sqrt{5}\approx 4.47$ s

(b) $v=u+gt=0+10(4.47)=44.7$ m/s

Common Mistakes

1. Confusing distance with displacement: Distance is scalar (path length); displacement is vector (shortest path).
2. Using average of speeds for average speed: Only works when time intervals are equal. Must use $\frac{\text{total distance}}{\text{total time}}$.
3. Forgetting sign conventions: Choose a positive direction consistently — velocity, acceleration, and displacement signs must follow the chosen convention.
4. Missing units in final answers: Always include proper SI units.
5. Treating velocity as speed: Speed = |velocity|. In 1D, velocity can be negative; speed cannot.

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