Kinematics Previous Year Questions — JEE Main Physics
38 past-year Kinematics questions from JEE Main (Physics) — page 3.
Q11 (2025)
The displacement $x$ versus time graph is shown below. {{IMG1}}
(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:
- (A), (D), (E) only
- (B), (C), (D) only
- (B), (D), (E) only
- (B), (C), (E) only
### Related Formula
$$\langle v \rangle = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$$
$$v_{\text{inst}} = \frac{dx}{dt} = \text{slope of } x\text{-}t \text{ graph}$$
### Core Logic
Let's test each statement using coordinates from the given graph:
- For (A): At $t=0, x=0$; at $t=3, x=5$. $\langle v \rangle = \frac{5-0}{3} = \frac{5}{3}\text{ m/s} \neq 10\text{ m/s}$ (Incorrect).
- For (B): At $t=3, x=5$; at $t=5, x=5$. $\langle v \rangle = \frac{5-5}{2} = 0\text{ m/s}$ (Correct).
### Step 1: Evaluate Remaining Statements
- 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: $\text{slope} = \frac{5 - (-10)}{3} = 5\text{ m/s}$. Thus, instantaneous velocity at $t=2\text{ s}$ is $5\text{ m/s}$ (Correct).
- For (D): Slope during 5 to 7s vs instantaneous slope at $t=6.5\text{ s}$ 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 $t=9\text{ s}$ is zero (Correct).
Thus, (B), (C), and (E) are the correct statements.
### Pattern Recognition
Average velocity requires only initial and final positions ($x_f, x_i$). Instantaneous velocity reads directly off the segment's geometric slope. If initial and final coordinates match, average velocity is unconditionally zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Q18 (2025)
A helicopter flying horizontally with a speed of $360~km/h$ at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point O, 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is:
(use acceleration due to gravity $g=10~m/s^{2}$ and neglect air resistance) [cite: 157, 158, 159, 160]
- $2\sqrt{5}~\mathrm{km}$ [cite: 161]
- $4~\mathrm{km}$ [cite: 163]
- $7.2~\mathrm{km}$ [cite: 162]
- $2\sqrt{2}~\mathrm{km}$ [cite: 163]
### Related Formula
$x = u \cdot t$ [cite: 763]
$$H = \frac{1}{2}gt^2$$ [cite: 764]
$$D = \sqrt{x^2 + H^2}$$ [cite: 774]
### Core Logic
First, convert the horizontal velocity component to metric SI units: [cite: 157, 763]
$$u = 360 \times \frac{5}{18} = 100\ \text{m/s}$$ [cite: 157, 763]
Calculate the horizontal range distance $x$ covered over $t = 20\ \text{s}$: [cite: 158, 763]
$$x = 100 \times 20 = 2000\ \text{m} = 2\ \text{km}$$ [cite: 158, 763]
The vertical displacement height $H$ is explicitly given as $2\ \text{km} = 2000\ \text{m}$[cite: 157, 769]. Let's confirm with free-fall height calculation matching the solution template: [cite: 764]
$$H = \frac{1}{2} \times 10 \times (20)^2 = 5 \times 400 = 2000\ \text{m} = 2\ \text{km}$$ [cite: 158, 769]
Now find the net spatial vector displacement $D$ from the release coordinates: [cite: 774]
$$D = \sqrt{x^2 + H^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\ \text{km}$$ [cite: 774]
### Pattern Recognition
Be careful with the wording: the question asks for the displacement from the *release coordinate position* [cite: 159], which is the hypotenuse vector $\sqrt{x^2 + H^2}$[cite: 774]. Do not mistake it for the horizontal range distance alone.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Plane
Q8 (2025)
The position vector of a moving body at any instant of time is given as $\vec{\mathrm{r}} = (5\mathrm{t}^2\hat{\mathrm{i}} - 5\mathrm{t}\hat{\mathrm{j}})\mathrm{m}$. The magnitude and direction of velocity at $t = 2s$ is,
- $5\sqrt{15}$ m/s, making an angle of $\tan^{-1}4$ with -ve Y axis
- $5\sqrt{15}$ m/s, making an angle of $\tan^{-1}4$ with +ve X axis
- $5\sqrt{17}$ m/s, making an angle of $\tan^{-1}4$ with -ve Y axis
- $5\sqrt{17}$ m/s, making an angle of $\tan^{-1}4$ with +ve X axis
### Related Formula
$$\vec{v} = \frac{d\vec{r}}{dt}$$
### Core Logic
Given position vector:
$$\vec{r} = 5t^2\hat{i} - 5t\hat{j}$$
Differentiating with respect to $t$:
$$\vec{v} = 10t\hat{i} - 5\hat{j}$$
At $t = 2\ \mathrm{s}$:
$$\vec{v} = 20\hat{i} - 5\hat{j}$$
Magnitude of velocity:
$$v = \sqrt{(20)^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17}\ \mathrm{m/s}$$
Direction analysis: {{SOL_IMG1}}
$v_x = 20$ along $+x$ axis, $v_y = -5$ along $-y$ axis.
Let $\theta$ be the angle made with the negative $Y$-axis:
$$\tan \theta = \left|\frac{v_x}{v_y}\right| = \frac{20}{5} = 4 \implies \theta = \tan^{-1}4$$
### Pattern Recognition
Always read the reference axis carefully in direction questions. Here, the angle is measured from the negative $Y$-axis, making $\tan\theta = v_x / |v_y|$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Plane
Q9 (2025)
The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between $t = 0$ to $t = 4s$ ?
{{IMG1}}
- $30\mathrm{m}$
- $10\mathrm{m}$
- $13\mathrm{m}$
- $11\mathrm{m}$
### Related Formula
The distance traveled by an object equals the total area enclosed under its velocity-time ($v-t$) plot along the time axis, treating all regional boundaries as strictly positive metrics:
$$\text{Distance} = \int |v| \, dt$$
### Core Logic
From the geometric grid profile between $t = 0$ and $t = 4\text{ s}$:
1. **First Region ($t=0$ to $t=2\text{ s}$)**: Forms a \right-angled \triangle with base $= 2\text{ s}$ and peak height $= 10\text{ ms}^{-1}$.
$$\text{Area}_1 = \frac{1}{2} \times 2 \times 10 = 10 \text{ m}$$
2. **Second Region ($t=2$ to $t=4\text{ s}$)**: Forms a standard rectangle with width $= (4 - 2) = 2\text{ s}$ and height $= 10\text{ ms}^{-1}$.
$$\text{Area}_2 = 2 \times 10 = 20 \text{ m}$$
Summing the areas together to extract total displacement path:
$$\text{Total Distance} = 10 + 20 = 30 \text{ m}$$
### Pattern Recognition
Always differentiate between distance and displacement on graph tracks. Displacement treats components below the axis as negative fields, while distance calculates absolute geometric magnitudes without direction.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Q1277 (2025)
Two projectiles are fired with same initial speed from same point on ground at angles of $(45^{\circ} - \alpha)$ and $(45^{\circ} + \alpha)$ , respectively, with the horizontal direction. The ratio of their maximum heights attained is :
- $\frac{1 - \tan\alpha}{1 + \tan\alpha}$
- $\frac{1 + \sin \alpha}{1 - \sin \alpha}$
- $\frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$
- $\frac{1 + \sin 2\alpha}{1 - \sin 2\alpha}$
### Related Formula
$$H_{\max} = \frac{u^2 \sin^2 \theta}{2g}$$
### Core Logic
Let $\theta_1 = 45^{\circ} - \alpha$ and $\theta_2 = 45^{\circ} + \alpha$. The ratio of maximum heights is:
$$\frac{H_1}{H_2} = \frac{\sin^2(45^{\circ} - \alpha)}{\sin^2(45^{\circ} + \alpha)}$$
### Step 1: Simplify Trigonometric Terms
$$\sin(45^{\circ} \pm \alpha) = \frac{1}{\sqrt{2}}\cos\alpha \pm \frac{1}{\sqrt{2}}\sin\alpha$$
$$\frac{H_1}{H_2} = \frac{(\cos\alpha - \sin\alpha)^2}{(\cos\alpha + \sin\alpha)^2} = \frac{\cos^2\alpha + \sin^2\alpha - 2\sin\alpha\cos\alpha}{\cos^2\alpha + \sin^2\alpha + 2\sin\alpha\cos\alpha} = \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$$
### Pattern Recognition
For complementary angles shifted symmetric to $45^{\circ}$, ratios simplify cleanly via $\sin 2\alpha$ identities.
### Chapter Mix
Class 11 Physics: Motion in a Plane
