The displacement x versus time graph is shown below. (A) The average velocity during 0 to 3 s is 10 m/s (B) The average velocity during 3 to 5 s is 0 m/s (C) The instantaneous velocity at t=2 s is 5 m/s (D) The average velocity during 5 to 7 s and instantaneous velocity at t=6.5 s are equal (E) The average velocity from t=0 to t=9 s is zero Choose the correct answer from the options given below:

Let's test each statement using coordinates from the given graph: - For (A): At t=0, x=0; at t=3, x=5. v = (5-0/3) = (5/3) m/s neq 10 m/s (Incorrect). - For (B): At t=3, x=5; at t=5, x=5. v = (5-5/2) = 0 m/s (Correct). - For (C): Segment from 0 to 3s passes through points (0, -5) or starts linearly. The slope from t=0 to t=3 can be calculated from the linear line segment: slope = (5 - (-10)/3) = 5 m/s. Thus, instantaneous velocity at t=2 s is 5 m/s (Correct). - For (D): Slope during 5 to 7s vs instantaneous slope at t=6.5 s are completely different because the path changes slope. - For (E): At t=0, x=-5 and at t=9, x=-5. Since net displacement is zero, the average velocity from t=0 to t=9 s is zero (Correct). Thus, (B), (C), and (E) are the correct statements. Average velocity requires only initial and final positions (xf, xi). Instantaneous velocity reads directly off the segment's geometric slope. If initial and final coordinates match, average velocity is unconditionally zero. Class 11 Physics: Motion in a Straight Line

The displacement x versus time graph is shown below. (A) The average velocity during 0 to 3 s is 10 m/s (B) The average velocity during 3 to 5 s is 0 m/s (C) The instantaneous velocity at t=2 s is 5 m

Let's test each statement using coordinates from the given graph: - For (A): At t=0, x=0; at t=3, x=5. v = (5-0/3) = (5/3) m/s neq 10 m/s (Incorrect). - For (B): At t=3, x=5; at t=5, x=5. v = (5-5/2) = 0 m/s (Correct). - For (C): Segment from 0 to 3s passes through points (0, -5) or starts linearly. The slope from t=0 to t=3 can be calculated from the linear line segment: slope = (5 - (-10)/3) = 5 m/s. Thus, instantaneous velocity at t=2 s is 5 m/s (Correct). - For (D): Slope during 5 to 7s vs i

The displacement x versus time graph is shown below. (A) The average velocity during 0 to 3 s is 10 m/s (B) The average velocity during 3 to 5 s is 0 m/s (C) The instantaneous velocity at t=2 s is 5 m/s (D) The average velocity during 5 to 7 s and instantaneous velocity at t=6.5 s are equal (E) The average velocity from t=0 to t=9 s is zero Choose the correct answer from the options given below:

Let's test each statement using coordinates from the given graph: - For (A): At t=0, x=0; at t=3, x=5. v = (5-0/3) = (5/3) m/s neq 10 m/s (Incorrect). - For (B): At t=3, x=5; at t=5, x=5. v = (5-5/2) = 0 m/s (Correct). - For (C): Segment from 0 to 3s passes through points (0, -5) or starts linearly. The slope from t=0 to t=3 can be calculated from the linear line segment: slope = (5 - (-10)/3) = 5 m/s. Thus, instantaneous velocity at t=2 s is 5 m/s (Correct). - For (D): Slope during 5 to 7s vs instantaneous slope at t=6.5 s are completely different because the path changes slope. - For (E): At t=0, x=-5 and at t=9, x=-5. Since net displacement is zero, the average velocity from t=0 to t=9 s is zero (Correct). Thus, (B), (C), and (E) are the correct statements. Average velocity requires only initial and final positions (xf, xi). Instantaneous velocity reads directly off the segment's geometric slope. If initial and final coordinates match, average velocity is unconditionally zero. Class 11 Physics: Motion in a Straight Line

The displacement x versus time graph is shown below. (A) The average velocity during 0 to 3 s is 10 m/s (B) The average velocity during 3 to 5 s is 0 m/s (C) The instantaneous velocity at t=2 s is 5 m

Let's test each statement using coordinates from the given graph: - For (A): At t=0, x=0; at t=3, x=5. v = (5-0/3) = (5/3) m/s neq 10 m/s (Incorrect). - For (B): At t=3, x=5; at t=5, x=5. v = (5-5/2) = 0 m/s (Correct). - For (C): Segment from 0 to 3s passes through points (0, -5) or starts linearly. The slope from t=0 to t=3 can be calculated from the linear line segment: slope = (5 - (-10)/3) = 5 m/s. Thus, instantaneous velocity at t=2 s is 5 m/s (Correct). - For (D): Slope during 5 to 7s vs i

The displacement x versus time graph is shown below. (A) The average velocity during 0 to 3 s is 10 m/s (B) The average…

Solution & Explanation

Related Formula

v = (Δ x)/(Δ t) = (xf - xᵢ)/(tf - tᵢ)

vᵢₙₛₜ = (dx)/(dt) = slope of x-t graph

Core Logic

Let's test each statement using coordinates from the given graph:

For (A): At $t=0, x=0$ $t=0, x=0$; at $t=3, x=5$ $t=3, x=5$. $v = (5-0)/(3) = (5)/(3) m/s \neq 10 m/s$ (Incorrect).
For (B): At $t=3, x=5$ $t=3, x=5$; at $t=5, x=5$ $t=5, x=5$. $v = (5-5)/(2) = 0 m/s$ (Correct).

Step 1: Evaluate Remaining Statements

For (C): Segment from 0 to 3s passes through points $(0, -5)$ $(0, -5)$ or starts linearly. The slope from $t=0$ $t=0$ to $t=3$ $t=3$ can be calculated from the linear line segment: $slope = (5 - (-10))/(3) = 5 m/s$ . Thus, instantaneous velocity at $t=2 s$ $t=2\text{ s}$ is $5 m/s$ $5\text{ m/s}$ (Correct).
For (D): Slope during 5 to 7s vs instantaneous slope at $t=6.5 s$ $t=6.5\text{ s}$ are completely different because the path changes slope.
For (E): At $t=0, x=-5$ $t=0, x=-5$ and at $t=9, x=-5$ $t=9, x=-5$. Since net displacement is zero, the average velocity from $t=0$ $t=0$ to $t=9 s$ $t=9\text{ s}$ is zero (Correct).

Thus, (B), (C), and (E) are the correct statements.

Pattern Recognition

Average velocity requires only initial and final positions ( $xf, xᵢ$ $x_f, x_i$). Instantaneous velocity reads directly off the segment's geometric slope. If initial and final coordinates match, average velocity is unconditionally zero.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

More Motion in a Straight Line 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: Evaluate Remaining Statements

Pattern Recognition

Chapter Mix

Reference Study Guides

More Motion in a Straight Line 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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Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Remaining Statements

Pattern Recognition

Chapter Mix

Reference Study Guides

More Motion in a Straight Line 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

More Motion in a Straight Line Questions — jee_main_2025_04_april_evening

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