Kinematics Previous Year Questions — JEE Main Physics
38 past-year Kinematics questions from JEE Main (Physics) — page 4.
Q1300 (2025)
The maximum speed of a boat in still water is $27 \, \text{km/h}$ . Now this boat is moving downstream in a river flowing at $9 \, \text{km/h}$ . A man in the boat throws a ball vertically upwards with speed of $10 \, \text{m/s}$ . Range of the ball as observed by an observer at rest on the river bank, is ________ cm. (Take $g = 10 \, \text{m/s}^2$ )
### Related Formula
$$T = \frac{2u_y}{g}, \quad R = v_x \cdot T$$
### Core Logic
{{SOL_IMG1}}
The total horizontal velocity components combine due to downstream motion addition
$$v_x = 27 + 9 = 36\text{ km/h} = 36 \cdot \frac{5}{18} = 10\text{ m/s}$$
The time of flight for the vertical launch tracking stands at :
$$T = \frac{2 \cdot 10}{10} = 2\text{ s}$$
Horizontal range measured along the river bank frame is :
$$Range = v_x \cdot T = 10 \cdot 2 = 20\text{ m} = 2000\text{ cm}$$
### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Class 11 Physics: Motion in a Plane
Q47 (2024)
A particle moving in a circle of radius $R$ with uniform speed takes time $T$ to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4R$. The angle of projection $\theta$ is then given by:
- $\sin^{-1}\left[\frac{2\mathrm{gT}^2}{pi^2\mathrm{R}}\right]^{\frac{1}{2}}$
- $\sin^{-1}\left[\frac{\pi^2\mathbf{R}}{2\mathrm{gT}^2}\right]^{\frac{1}{2}}$
- $\cos^{-1}\left[\frac{2\mathrm{gT}^2}{pi^2\mathrm{R}}\right]^{\frac{1}{2}}$
- $\cos^{-1}\left[\frac{\pi R}{2gT^2}\right]^{\frac{1}{2}}$
### Related Formula
Uniform circular velocity:
$$v = \frac{2\pi R}{T}$$
Maximum height of a projectile:
$$H = \frac{v^2 \sin^2\theta}{2g}$$
### Core Logic
Given that the projectile max height matches $H = 4R$:
$$4R = \frac{v^2 \sin^2\theta}{2g}$$
Substitute the value of $v$ from the circular motion loop:
$$4R = \frac{\left(\frac{2\pi R}{T}\right)^2 \sin^2\theta}{2g}$$
$$4R = \frac{4\pi^2 R^2 \sin^2\theta}{2g T^2}$$
### Step 1: Isolate Angular Components
Cancel out $4R$ from both sides:
$$1 = \frac{\pi^2 R \sin^2\theta}{2g T^2}$$
$$\sin^2\theta = \frac{2g T^2}{\pi^2 R}$$
$$\sin\theta = \left(\frac{2g T^2}{\pi^2 R}\right)^{\frac{1}{2}} \implies \theta = \sin^{-1}\left[\frac{2gT^2}{\pi^2 R}\right]^{\frac{1}{2}}$$
### Pattern Recognition
Connect circular metrics directly to projectile parameters via velocity matching. Keeping terms unsimplified makes it easy to cancel common elements later.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Plane
Q60 (2024)
A particle is moving in one dimension (along x-axis) under the action of a variable force. It's initial position was $16\mathrm{~m}$ right of origin. The variation of its position ($x$) with time ($t$) is given as $x = -3t^3 + 18t^2 + 16t$ where $x$ is in m and $t$ is in s. The velocity of the particle when its acceleration becomes zero is _______ $\mathrm{m/s}$.
### Related Formula
Velocity via first derivative of position:
$$v = \frac{dx}{dt}$$
Acceleration via derivative of velocity:
$$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$$
### Core Logic
Given $x(t) = -3t^3 + 18t^2 + 16t$.
Differentiate once to find velocity $v(t)$:
$$v = \frac{dx}{dt} = -9t^2 + 36t + 16$$
Differentiate again to find acceleration $a(t)$:
$$a = \frac{dv}{dt} = -18t + 36$$
### Step 1: Find the Time When Acceleration is Zero
Set the acceleration equation to zero:
$$-18t + 36 = 0 \implies 18t = 36 \implies t = 2\mathrm{~s}$$
### Step 2: Calculate Velocity at Target Time
Substitute $t = 2\mathrm{~s}$ back into the velocity function:
$$v(2) = -9(2)^2 + 36(2) + 16$$
$$v(2) = -36 + 72 + 16 = 36 + 16 = 52\mathrm{~ms}^{-1}$$
*Note on original structural text step matching:* The original calculation step text features a minor print truncation ($v = -9t^2 + 36 + 16$), but explicitly evaluates out to the correct total of $52\mathrm{~m/s}$.
### Pattern Recognition
Inflection point tracking: Finding where acceleration equals zero is mathematically identical to finding the maximum velocity point on the curve.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Q47 (2024)
A particle is moving in a straight line. The variation of position $x$ as a function of time $t$ is given as $x = (t^3 - 6t^2 + 20t + 15)\text{ m}$. The velocity of the body when its acceleration becomes zero is:
- $4\text{ m/s}$
- $8\text{ m/s}$
- $10\text{ m/s}$
- $6\text{ m/s}$
### Related Formula
The relationship between position $x$, velocity $v$, and acceleration $a$ is given by differentiation with respect to time $t$:
$$v = \frac{dx}{dt}$$
$$a = \frac{dv}{dt}$$
### Core Logic
Given position:
$$x = t^3 - 6t^2 + 20t + 15$$
Differentiating once to find velocity $v$:
$$v = \frac{dx}{dt} = 3t^2 - 12t + 20$$
Differentiating again to find acceleration $a$:
$$a = \frac{dv}{dt} = 6t - 12$$
### Step 1: Determine Time when Acceleration is Zero
Set the acceleration to zero:
$$a = 0 \implies 6t - 12 = 0 \implies t = 2\text{ seconds}$$
So, the acceleration becomes zero at $t = 2\text{ s}$.
### Step 2: Calculate Velocity at this Time
Substitute $t = 2\text{ s}$ into the velocity equation:
$$v = 3(2)^2 - 12(2) + 20$$
$$v = 3(4) - 24 + 20$$
$$v = 12 - 24 + 20 = 8\text{ m/s}$$
Thus, the velocity is $8\text{ m/s}$.
### Pattern Recognition
Sees: displacement function of degree 3 → acceleration is linear in time. The zero of a linear function of form $At - B = 0$ is easily calculated, and substituting back resolves to a basic quadratic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Q31 (2024)
Position of an ant ($S$ in metres) moving in Y-Z plane is given by $\vec{S} = 2t^{2}\hat{j} + 5\hat{k}$ (where $t$ is in seconds).
The magnitude and direction of velocity of the ant at $t = 1$ s will be:
- $16\text{ m/s in y-direction}$
- $4\text{ m/s in x-direction}$
- $9\text{ m/s in z-direction}$
- $4\text{ m/s in y-direction}$
### Related Formula
$$\vec{v} = \frac{d\vec{S}}{dt}$$
### Core Logic
Differentiating the position vector with respect to time gives the velocity vector:
$$\vec{v} = \frac{d}{dt}(2t^{2}\hat{j} + 5\hat{k}) = 4t\hat{j}$$
### Step 1: Evaluation at given time
Substitute $t = 1\text{ s}$ into the velocity expression:
$$\vec{v} = 4(1)\hat{j} = 4\hat{j}\text{ m/s}$$
### Pattern Recognition
Constant term vectors (like $5\hat{k}$) drop out during differentiation. The resulting velocity only has a component along the $\hat{j}$ direction, confirming motion purely parallel to the y-axis at that instant.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
