Class 11 Physics: Motion in a Straight Line | Physics Q&A

CBSE Class 11 › Unit II: Kinematics

Motion in a Straight Line

NCERT Chapter 2 • Velocity, Acceleration & Kinematic Equations

NCERT 2025–26 Unit II • ~8 Marks JEE Main • 1-2 Questions

High-speed train on a straight track with kinematic graphs. — Rectilinear Motion: The physics of moving in a straight line.

1. Path Length and Displacement

Motion is the change in position of an object with time. To describe this, we use a reference point (origin).

Path Length (Distance): The total length of the actual path traversed by the object. It is a scalar quantity (always positive).
Displacement ( $\Delta x$ ): The shortest distance between the initial and final positions. It is a vector quantity (can be positive, negative, or zero).

Diagram comparing total path length and displacement vector. — Distance is the actual path; Displacement is the shortest route.

$\Delta x = x_2 - x_1$

Where $x_2$ is final position and $x_1$ is initial position.

Problem 1: A person moves in a semi-circle of radius $R$ from point A to B. Find distance and displacement.

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Distance = Circumference/2 = $\pi R$ .
Displacement = Diameter = $2R$ .

Problem 2: Can displacement be greater than path length?

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No. Displacement is the shortest path. It can be equal to path length (straight line motion without turning back) or less than it, but never greater.

2. Average Velocity and Speed

Average Velocity ( $\bar{v}$ )

Defined as the change in position or displacement divided by the time interval.

$\bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_2 - x_1}{t_2 - t_1}$

Average Speed

Defined as the total path length travelled divided by the total time interval.

$\text{Average Speed} = \dfrac{\text{Total Path Length}}{\text{Total Time}}$

Instantaneous Velocity ( $v$ ):

The velocity at a specific instant of time. It is defined as the limit of average velocity as the time interval $\Delta t$ approaches zero.
$v = \lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}$

Problem 3: A car travels from A to B at 40 km/h and returns from B to A at 60 km/h. Find average speed and average velocity.

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Average Velocity: 0 (Since displacement is zero).
Average Speed: Harmonic Mean = $\dfrac{2v_1v_2}{v_1+v_2} = \dfrac{2(40)(60)}{40+60} = \dfrac{4800}{100} = 48$ km/h.

3. Acceleration

Acceleration is the rate of change of velocity with time.

$a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$

x-t graphs showing stationary, uniform motion, and accelerated motion. — Interpreting Position-Time (x-t) Graphs.

Understanding Signs:
* Positive Acceleration: Velocity increases in positive direction.
* Negative Acceleration (Retardation): Velocity decreases (if moving positively).

Problem 4: If $x = t^3 - 6t^2$ , find the time when acceleration is zero.

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Velocity $v = \dfrac{dx}{dt} = 3t^2 - 12t$ .
Acceleration $a = \dfrac{dv}{dt} = 6t - 12$ .
Set $a = 0 \Rightarrow 6t = 12 \Rightarrow t = 2$ s.

4. Kinematic Equations (Uniform Acceleration)

For objects moving with constant acceleration ( $a$ ), we use three key equations. While these can be derived graphically, the calculus method is essential for competitive exams.

Check out our detailed guide on how easy is to derive the 5 equations of motion

1. $\quad v = v_0 + at$

2. $\quad x = v_0t + \dfrac{1}{2}at^2$

3. $\quad v^2 = v_0^2 + 2ax$

Velocity-time graph showing area under the curve representing displacement. — The area under a v-t graph gives the displacement $\Delta x$ .

Derivation 1:

Velocity-Time Relation ( $v = v_0 + at$ )

Calculus Method

Step 1: Definition of Acceleration

$a = \dfrac{dv}{dt} \Rightarrow dv = a dt$

Step 2: Integration
Integrate both sides with limits ( $v_0$ to $v$ ) and ( $0$ to $t$ ):

$\int_{v_0}^{v} dv = \int_{0}^{t} a dt$

Step 3: Solve
Since $a$ is constant, take it out of integral:

$[v]_{v_0}^{v} = a [t]_{0}^{t}$

$v - v_0 = a(t - 0)$

$v = v_0 + at$

Derivation 2:

Position-Time Relation ( $x = v_0t + \frac{1}{2}at^2$ )

Calculus Method

Step 1: Definition of Velocity

$v = \dfrac{dx}{dt} \Rightarrow dx = v dt$

Step 2: Substitute $v$
From the first equation, $v = v_0 + at$ .

$dx = (v_0 + at) dt$

Step 3: Integration
Integrate from initial position $x_0$ (usually 0) to final $x$ :

$\int_{0}^{x} dx = \int_{0}^{t} (v_0 + at) dt$

$x = \int_{0}^{t} v_0 dt + \int_{0}^{t} at dt$

Step 4: Solve

$x = v_0 [t]_{0}^{t} + a \left[\dfrac{t^2}{2}\right]_{0}^{t}$

$x = v_0t + \dfrac{1}{2}at^2$

Derivation 3:

Velocity-Position Relation ( $v^2 = v_0^2 + 2ax$ )

Calculus Method

Step 1: Chain Rule for Acceleration

$a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = v \dfrac{dv}{dx}$

$a dx = v dv$

Step 2: Integration
Integrate with position limits ( $0$ to $x$ ) and velocity limits ( $v_0$ to $v$ ):

$\int_{0}^{x} a dx = \int_{v_0}^{v} v dv$

Step 3: Solve

$a [x]_{0}^{x} = \left[\dfrac{v^2}{2}\right]_{v_0}^{v}$

$ax = \dfrac{v^2}{2} - \dfrac{v_0^2}{2}$

$2ax = v^2 - v_0^2$

$v^2 = v_0^2 + 2ax$

Problem 5: A ball is from a tower of height 20m. Find the time taken to reach the ground. ( $g=10$ m/s²)

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Using $x = v_0t + \dfrac{1}{2}at^2$ :
$20 = 0(t) + \dfrac{1}{2}(10)t^2$
$20 = 5t^2 \Rightarrow t^2 = 4 \Rightarrow t = 2$ s.

Problem 6: A car moving at $u$ stops after distance $d$ upon braking. If speed is doubled ( $2u$ ), what is the new stopping distance?

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Using $v^2 = u^2 + 2ax$ , with $v=0$ :
$0 = u^2 - 2ad \Rightarrow d = \dfrac{u^2}{2a}$ .
Stopping distance is proportional to $u^2$ .
If $u$ becomes $2u$ , $d$ becomes $(2)^2 = 4$ times.
New distance = $4d$ .

5. Relative Velocity

The velocity of object A with respect to object B is given by:

$v_{AB} = v_A - v_B$

Illustration of relative velocity for objects moving in parallel and anti-parallel directions. — Relative Velocity depends on direction: Subtract for same direction, Add for opposite.

Problem 7: Two trains A and B of length 100m each are moving in opposite directions with 36 km/h and 54 km/h. Find time to cross each other.

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$v_A = 36$ km/h = 10 m/s.
$v_B = -54$ km/h = -15 m/s.
Relative Velocity $v_{AB} = v_A - v_B = 10 - (-15) = 25$ m/s.
Total distance = $L_A + L_B = 100 + 100 = 200$ m.
Time $t = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{200}{25} = 8$ s.

You can also solve this problem on relative velocity.

← Previous Chapter Next Chapter: Motion in a Plane →

1. Path Length and Displacement

2. Average Velocity and Speed

Average Velocity ()

Average Speed

3. Acceleration

4. Kinematic Equations (Uniform Acceleration)

5. Relative Velocity

Average Velocity ( $\bar{v}$ )