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…

Q25 jee_main_2025_02_april_morning Motion in a Straight Line

A person travelling on a straight line moves with a uniform velocity

v₁

for a distance

x

and with a uniform velocity

v₂

for the next

(3)/(2)x

distance. The average velocity in this motion is

(50)/(7)~m/s

. If

v₁

is

5~m/s

then

v₂ =

____

m/s

.

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

vavg = Total DistanceTotal Time

Core Logic

Let's find the time taken for each section of the motion:

First section of distance $x$ $x$ with velocity $v₁ = 5~m/s$ $v_1 = 5\mathrm{~m/s}$:

t₁ = (x)/(v₁) = (x)/(5)

Second section of distance $(3)/(2)x$ $\frac{3}{2}x$ with velocity $v₂$ $v_2$:

t₂ = (3x/2)/(v₂) = (3x)/(2v₂)

Total distance is:

dtotal = x + (3)/(2)x = (5)/(2)x

Average velocity is:

vavg = dtotalt₁ + t₂ = ((5)/(2)x)/((x)/(5) + (3x)/(2v₂))

We are given $vavg = (50)/(7)~m/s$ . Equating the two values (noting $x$ $x$ cancels out):

(50)/(7) = ((5)/(2))/((1)/(5) + (3)/(2v₂))

Divide both sides by 5:

(10)/(7) = ((1)/(2))/((1)/(5) + (3)/(2v₂)) 10 ((1)/(5) + (3)/(2v₂)) = (7)/(2)

2 + (15)/(v₂) = 3.5 (15)/(v₂) = 1.5 v₂ = (15)/(1.5) = 10~m/s

Step 1: Final Conclusion

The velocity $v₂$ $v_2$ is $10~m/s$ $10\mathrm{~m/s}$.

Pattern Recognition

Never take simple arithmetic averages of velocities! Average velocity must always be calculated as $Total DistanceTotal Time$ . Because total distance and time intervals are proportional to $x$ $x$, $x$ $x$ cleanly cancels out.

Chapter Mix

Class 11 Physics: Kinematics

Q11 jee_main_2025_03_april_evening Projectile Motion

A particle is projected with velocity

u

so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as

(nu²)/(25g)

, where value of

n

is: (Given '

g

' is the acceleration due to gravity).

A. 6
B. 18
C. 12
D. 24

Solution

Related Formula

For a projectile with launch speed $u$ $u$ and angle $θ$ $\theta$:

Horizontal Range:

R = (u² (2θ))/(g) = (2 u² θ θ)/(g)

Maximum Height:

H = (u² ²θ)/(2g)

The general ratio linking range and maximum height is:

θ = (4H)/(R)

Core Logic

Given state:

R = 3H \Rightarrow (H)/(R) = (1)/(3)

Step 1: Determine the projection angle ( $θ$ $\theta$)

Substitute the ratio into the relation:

θ = 4 ((H)/(R)) = 4 ((1)/(3)) = (4)/(3)

This is a standard Pythagorean triangle angle:

θ = (4)/(5), θ = (3)/(5)

Step 2: Compute the horizontal range ( $R$ $R$)

R = (2 u² θ θ)/(g)

R = (2 u² ((4)/(5)) ((3)/(5)))/(g) = (24 u²)/(25 g)

Comparing this with the given format $(nu²)/(25g)$ $\frac{nu^2}{25g}$:

$n = 24$ $n = 24$

Pattern Recognition

The relation $θ = 4H/R$ $\tan\theta = 4H/R$ is an essential identity in projectile dynamics. Whenever $R = k H$ $R = k H$, then $θ = 4/k$ $\tan\theta = 4/k$. Recognizing standard angles like $θ = 4/3$ $\tan\theta = 4/3$ or $3/4$ $3/4$ directly yields trigonometric values immediately.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q14 jee_main_2025_03_april_evening Kinematics and Derivative Relations

A particle moves along the x-axis and has its displacement x varying with time t according to the equation

x=c₀(t²-2)+c(t-2)²

where

c₀

and c are constants of appropriate dimensions. Then, which of the following statements is correct?

A. the acceleration of the particle is $2c₀$ $2c_{0}$
B. the acceleration of the particle is 2c
C. the initial velocity of the particle is 4c
D. the acceleration of the particle is $2(c+c₀)$ $2(c+c_{0})$

Solution

Related Formula

In rectilinear kinematics:

Velocity:

v = (dx)/(dt)

Acceleration:

a = (dv)/(dt) = (d²x)/(dt²)

Core Logic Given position-time function:

x(t) = c₀ (t² - 2) + c (t - 2)²

Step 1: Differentiate once to get velocity ($ Step 1: Differentiate once to get velocity ($v) $) $

v = (dx)/(dt) = (d)/(dt)[c₀(t² - 2)] + (d)/(dt)[c(t-2)²]

v = c₀ (2t) + c \cdot 2(t-2) = 2 c₀ t + 2 c(t - 2)

Step 2: Differentiate again to get acceleration ($ Step 2: Differentiate again to get acceleration ($a) $) $

a = (dv)/(dt) = (d)/(dt)[2 c₀ t + 2 c(t - 2)]

a = 2 c₀ + 2 c = 2(c + c₀)

This shows acceleration is constant and equals$ This shows acceleration is constant and equals $2(c + c_0), matching Statement (4). Pattern Recognition Whenever a position function is a pure quadratic polynomial in$, matching Statement (4). Pattern Recognition Whenever a position function is a pure quadratic polynomial in $t, the acceleration is constant and equal to$, the acceleration is constant and equal to $2 \timesthe coefficient of the$ the coefficient of the $t^2term. Rewriting$ term. Rewriting $x(t):

x(t) = (c₀ + c)t² - 4ct + (4c - 2c₀)

The coefficient of $ The coefficient of $t^2 is $ is $ (c_0 + c) . Thus, acceleration is $. Thus, acceleration is $ 2(c_0 + c)$ directly. Chapter Mix Class 11 Physics: Motion in a Straight Line

Q12 jee_main_2025_07_april_morning Projectile Motion

Two projectiles are fired from ground with same initial speeds from same point at angles

(45° + α)

and

(45° - α)

with horizontal direction. The ratio of their times of flights is

A. 1
B. $(1 - α)/(1 + α)$ $\frac{1 - \tan\alpha}{1 + \tan\alpha}$
C. $(1 + 2α)/(1 - 2α)$
D. $(1 + α)/(1 - α)$ $\frac{1 + \tan\alpha}{1 - \tan\alpha}$

Solution

Related Formula

The time of flight $T$ $T$ of a projectile launched with speed $u$ $u$ at an angle $θ$ $\theta$ with the horizontal is:

T = (2u θ)/(g)

Core Logic The launch angles of the two projectiles are:

θ₁ = 45^° + α

θ₂ = 45^° - α

Since they have the same speed $ Since they have the same speed $u :

(T₁)/(T₂) = ( (45^° + α))/( (45^° - α))

Step 1: Simplify Trigonometric Ratio Using the angle sum and difference formulas:

(T₁)/(T₂) = ( 45^° α + 45^° α)/( 45^° α - 45^° α)

(T₁)/(T₂) = 1\sqrt(2) α + 1\sqrt(2) α 1\sqrt(2) α - 1\sqrt(2) α = ( α + α)/( α - α)

Divide numerator and denominator by$ Divide numerator and denominator by $\cos\alpha:

(T₁)/(T₂) = (1 + α)/(1 - α)

$

Pattern Recognition Sees: Projectile angles complementary to$ Pattern Recognition Sees: Projectile angles complementary to $45^\circ. Shortcut: Remember the identity$. Shortcut: Remember the identity $\tan(45^\circ + \alpha) = \frac{1+\tan\alpha}{1-\tan\alpha}. Since complementary angles have sine ratios proportional to$. Since complementary angles have sine ratios proportional to $\sin(45^\circ + \alpha)/\sin(45^\circ - \alpha) = \tan(45^\circ + \alpha), the answer is directly$, the answer is directly $\frac{1+\tan\alpha}{1-\tan\alpha}$. Chapter Mix Class 11 Physics: Motion in a Plane

Q16 jee_main_2025_08_april_evening Projectile Motion

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is

8

times higher than that of the second ball.

T₁

and

T₂

are the total flying times of first and second ball, respectively, then the ratio of

T₁

and

T₂

is:

A. $2\sqrt(2) : 1$ $2\sqrt{2} : 1$
B. $2 : 1$ $2 : 1$
C. $\sqrt(2) : 1$ $\sqrt{2} : 1$
D. $4 : 1$ $4 : 1$

Solution

Related Formula

H = (u² ²θ)/(2g) and T = (2u θ)/(g)

where, $H$ $H$ = maximum height reached $T$ $T$ = total time of flight $u$ $u$ = initial projection velocity $θ$ $\theta$ = angle of projection

Core Logic

From the formulas, we see that:

H \propto ²θ and T \propto θ

Thus, we can directly link time of flight to the square root of the maximum height:

T \propto \sqrt(H) (T₁)/(T₂) = \sqrt((H₁)/(H₂))

Step 1: Compute Ratio

Given:

H₁ = 8 H₂ (H₁)/(H₂) = 8

Substitute this ratio:

(T₁)/(T₂) = \sqrt(8) = 2\sqrt(2)

Thus, the ratio is $2\sqrt(2) : 1$ $2\sqrt{2} : 1$.

Pattern Recognition

Sees: Projectile heights ratio → Time of flight ratio. Shortcut: Since $H \propto uy²$ $H \propto u_y^2$ and $T \propto uy$ $T \propto u_y$, we have $T \propto \sqrt(H)$ $T \propto \sqrt{H}$. If the height is 8 times larger, the flying time is $\sqrt(8) = 2\sqrt(2)$ $\sqrt{8} = 2\sqrt{2}$ times larger. ✓

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 3

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine the projection angle (θ$\theta$)

Step 2: Compute the horizontal range (R$R$)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Differentiate once to get velocity ($ Step 1: Differentiate once to get velocity ($v)$)

Step 1: Differentiate once to get velocity ($v)

Step 2: Differentiate again to get acceleration ($ Step 2: Differentiate again to get acceleration ($a)$)

Step 2: Differentiate again to get acceleration ($a)

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Simplify Trigonometric Ratio

Step 1: Simplify Trigonometric Ratio

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Ratio

Pattern Recognition

Chapter Mix

More Motion in a Straight Line Questions — jee_main_2025_04_april_evening

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Step 1: Determine the projection angle ( $θ$ $\theta$)

Step 2: Compute the horizontal range ( $R$ $R$)

Step 1: Differentiate once to get velocity ($)

Step 1: Differentiate once to get velocity ($v $)$

Step 2: Differentiate again to get acceleration ($)

Step 2: Differentiate again to get acceleration ($a $)$