Which of the following curves possibly represent one-dimensional motion of a particle? (A) (B) (C) (D) Choose the…

Q: Which of the following curves possibly represent one-dimensional motion of a particle? (A) (B) (C) (D) Choose the correct answer from the options given below:

### Related Formula For realistic physical motion in one dimension: 1. Time t can never be negative during a normal positive time sequence, and cannot flow backwards. 2. Total distance covered can never decrease over time. 3. A particle cannot have two different values of position or velocity at the exact same instant of time. ### Core Logic Let us analyze each curve: - **Curve (A)** (Phase phi vs Time t): Represents phi = kt + C, which is a valid linear relationship of phase over time (e.g., in

Year	2026	2025	2024	Total
Questions	8	16	14	38

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

For realistic physical motion in one dimension:

Time $t$ $t$ can never be negative during a normal positive time sequence, and cannot flow backwards.
Total distance covered can never decrease over time.
A particle cannot have two different values of position or velocity at the exact same instant of time.

Core Logic

Let us analyze each curve:

Curve (A) (Phase $φ$ $\phi$ vs Time $t$ $t$): Represents $φ = kt + C$ $\phi = kt + C$, which is a valid linear relationship of phase over time (e.g., in Simple Harmonic Motion $x = A (kt + C)$ $x = A\sin(kt + C)$). (Valid)
Curve (B) (Velocity $v$ $v$ vs Displacement $x$ $x$): A closed loop, which represents symmetric harmonic-type oscillation. For example, $v² + ω² x² = const$ $v^2 + \omega^2 x^2 = \text{const}$ (ellipse) is a perfectly physically valid 1D SHM velocity-displacement phase portrait. (Valid)
Curve (C) (Velocity vs Time): The curve enters into the negative time quadrant. Time cannot go backwards or exist in negative values relative to starting sequence in standard physical scenarios. (Invalid)
Curve (D) (Total Distance vs Time): Represents total distance increasing over time. Total distance is a non-decreasing function of time ( $d(d)/dt \geq 0$ $d(d)/dt \ge 0$). Thus, this curve is physically valid. (Valid)

Step 1: Conclusion

Therefore, curves A, B, and D possibly represent physical one-dimensional motion. The correct option is (1).

Pattern Recognition

Quick check for graph validity:

Time cannot run backwards (ruling out C).
Total distance can never decrease (D is valid because it strictly goes upwards).
$v$ $v$ vs $x$ $x$ can be circular/elliptical in SHM (B is valid).

Chapter Mix

Class 11 Physics: Motion in a Straight Line

More Kinematics Previous-Year Questions — Page 8

Q44 jee_main_2024_31_jan_evening Vector Algebra

If two vectors

A

and

B

having equal magnitude

R

are inclined at an angle

θ

, then

A. $| A - B| = \sqrt(2)R ((θ)/(2))$
B. $| A + B| = 2R ((θ)/(2))$
C. $| A + B| = 2R ((θ)/(2))$
D. $| A - B| = 2R ((θ)/(2))$

Solution

Related Formula

The magnitude of the resultant vector is given by:

| Rᵣₑₛ| = \sqrt(A² + B² + 2AB θ)

Core Logic

Let $| A| = | B| = R$ $|\vec{A}| = |\vec{B}| = R$. Then for vector addition:

| A + B| = \sqrt(R² + R² + 2R² θ)

Step 1: Simplify Addition Form

| A + B| = \sqrt(2R² (1 + θ))

Using the trigonometric identity $1 + θ = 2 ² ((θ)/(2))$ :

| A + B| = \sqrt(2R² \times 2 ² ((θ)/(2))) = 2R ((θ)/(2))

Step 2: Cross-check Subtraction Form

For subtraction:

| A - B| = \sqrt(R² + R² - 2R² θ)

| A - B| = \sqrt(2R² (1 - θ)) = \sqrt(2R² \times 2 ² ((θ)/(2))) = 2R ((θ)/(2))

Checking options, only $| A + B| = 2R ((θ)/(2))$ is correctly paired in the choice list.

Pattern Recognition

Standard geometry shortcut: Addition of two equal vectors yields a cosine half-angle dependency. Subtraction yields a sine half-angle dependency. $(+ \to )$ $(+ \to \cos)$, $(- \to )$ $(- \to \sin)$.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q jee_main_2024_31_jan_morning Projectile Motion

A body starts falling freely from height

H

hits an inclined plane in its path at height

h

. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of

(H)/(h)

for which the body will take the maximum time to reach the ground is

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

t = \sqrt((2d)/(g))

Core Logic

Projectile Motion diagram for Q55 - JEE Main 2024 Morning

The body falls freely from height $H$ $H$ to height $h$ $h$. The distance fallen is $(H - h)$ $(H - h)$. Time taken to fall this distance:

t₁ = \sqrt((2(H - h))/(g))

After elastic impact, the vertical velocity becomes zero (the entire velocity is directed horizontally). From height $h$ $h$, it now acts as a horizontal projectile. The time taken to reach the ground vertically from height $h$ $h$ is:

t₂ = \sqrt((2h)/(g))

Total time of flight $T = t₁ + t₂$ $T = t_1 + t_2$:

T = \sqrt((2(H - h))/(g)) + \sqrt((2h)/(g))

Step 2: Maximizing Time

To find the maximum time $T$ $T$, differentiate $T$ $T$ with respect to $h$ $h$ and equate to zero:

(dT)/(dh) = \sqrt((2)/(g)) ( -12\sqrt(H - h) + 12\sqrt(h) ) = 0

12\sqrt(h) = 12\sqrt(H - h)

\sqrt(H - h) = \sqrt(h)

Squaring both sides: $H - h = h$ $H - h = h$ $H = 2h$ $H = 2h$

(H)/(h) = 2

Chapter Mix

Class 11 Physics: Kinematics

Q33 jee_main_2024_31_jan_morning Differentiation In Kinematics

The relation between time '

t

' and distance '

x

' is

t = α x² + β x

, where

α

and

β

are constants. The relation between acceleration (

a

) and velocity (

v

) is:

A. $a = -2α v³$ $a = -2\alpha v^3$
B. $a = -5α v⁵$ $a = -5\alpha v^5$
C. $a = -3α v²$ $a = -3\alpha v^2$
D. $a = -4α v⁴$ $a = -4\alpha v^4$

Solution

Related Formula

v = (dx)/(dt)

a = (dv)/(dt) = v(dv)/(dx)

Step 1: Differentiate with respect to time

Given the equation:

t = α x² + β x

Differentiating with respect to time $t$ $t$:

(dt)/(dt) = (d)/(dt)(α x² + β x)

1 = 2α x (dx)/(dt) + β (dx)/(dt)

1 = (2α x + β) v

v = (2α x + β)⁻¹

Step 2: Calculate Acceleration

Now, acceleration $a = (dv)/(dt)$ $a = \frac{dv}{dt}$. Differentiating $v$ $v$ with respect to time $t$ $t$:

a = (d)/(dt) [ (2α x + β)⁻¹ ]

a = -1(2α x + β)⁻² \cdot (d)/(dt)(2α x + β)

a = -(2α x + β)⁻² \cdot (2α) (dx)/(dt)

Substitute $v$ $v$ and $(2α x + β)⁻² = v²$ $(2\alpha x + \beta)^{-2} = v^2$:

a = -(v²) \cdot (2α) \cdot v

a = -2α v³

Pattern Recognition

Standard kinematic shortcut: Whenever $t = Ax² + Bx$ $t = Ax^2 + Bx$, $v = (2Ax+B)⁻¹$ $v = (2Ax+B)^{-1}$ and $a = -2A v³$ $a = -2A v^3$. Memorizing this directly saves derivation time during the exam.

Chapter Mix

Class 11 Physics: Kinematics

Solution & Explanation

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

More Kinematics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Simplify Addition Form

Step 2: Cross-check Subtraction Form

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Maximizing Time

Chapter Mix

Solution

Related Formula

Step 1: Differentiate with respect to time

Step 2: Calculate Acceleration

Pattern Recognition

Chapter Mix

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More Kinematics Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Simplify Addition Form

Step 2: Cross-check Subtraction Form

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Maximizing Time

Chapter Mix

Solution

Related Formula

Step 1: Differentiate with respect to time

Step 2: Calculate Acceleration

Pattern Recognition

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