Kinematics Previous Year Questions — JEE Main Physics
38 past-year Kinematics questions from JEE Main (Physics).
Q15 (2025)
A sportsman runs around a circular track of radius $r$ such that he traverses the path ABAB. The distance travelled and displacement, respectively, are
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- $2\mathrm{r}, 3\pi \mathrm{r}$
- $3\pi \mathbf{r},\pi \mathbf{r}$
- $\pi \mathrm{r}, 3\mathrm{r}$
- $3\pi \mathrm{r}, 2\mathrm{r}$
### Related Formula
1. Distance: Total actual path length covered.
2. Displacement: Shortest straight-line distance connecting the initial and final position.
3. Circumference of a complete circle = $2\pi r$
4. Semicircular arc length = $\pi r$
### Core Logic
The trajectory is defined by the sequence of points $A \to B \to A \to B$:
1. **Distance Travelled:**
- Segment 1 ($A \to B$): Semicircular path of length $\pi r$
- Segment 2 ($B \to A$): Semicircular path of length $\pi r$
- Segment 3 ($A \to B$): Semicircular path of length $\pi r$
- Total distance:
$$\text{Distance} = \pi r + \pi r + \pi r = 3\pi r$$
2. **Displacement:**
- Initial position: $A$
- Final position: $B$
- Since points $A$ and $B$ represent diametrically opposite positions on the circle, the shortest distance between them is equal to the diameter of the circle:
$$\text{Displacement} = 2r$$
### Step 1: Write result
The actual distance travelled is $3\pi r$ and the magnitude of displacement is $2r$.
### Pattern Recognition
Sees: Circular kinematics path tracing.
Trap: Accidentally substituting straight chords for the distance arcs, or assuming the loop returns fully to A (which would yield a zero displacement).
Shortcut: The path ends at B. Since 3 half-loops are made, distance = $3 \times \pi r = 3\pi r$. The direct path from start A to end B is just the diameter $2r$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Q81 (2025)
A river is flowing from west to east direction with speed of $9\mathrm{~km~h}^{-1}$. If a boat capable of moving at a maximum speed of $27\mathrm{~km~h}^{-1}$ in still water, crosses the river in half a minute, while moving with maximum speed at an angle of $150^{\circ}$ to direction of river flow, then the width of the river is:
- $300\mathrm{~m}$
- $112.5\mathrm{~m}$
- $75\mathrm{~m}$
- $112.5\times \sqrt{3}\mathrm{~m}$
### Related Formula
$$v_{\perp} = v_{br} \sin\theta$$
$$w = v_{\perp} \cdot t$$
### Core Logic
Let the West-to-East river flow direction be along the positive $x$-axis.
The boat's velocity relative to water is $v_{br} = 27\mathrm{~km/h}$ at an angle of $150^{\circ}$ with respect to the river flow (which is $30^{\circ}$ upstream from the perpendicular crossing line).
To find the width of the river, we only need the component of the boat's velocity perpendicular to the river bank ($y$-axis):
$$v_{\perp} = v_{br} \sin(150^{\circ}) = 27 \sin(150^{\circ}) = 27 \times \frac{1}{2} = 13.5\mathrm{~km/h}$$
Convert this velocity to SI units:
$$v_{\perp} = 13.5 \times \frac{5}{18}\mathrm{~m/s} = 3.75\mathrm{~m/s}$$
Given the crossing time is half a minute ($t = 30\mathrm{~s}$):
$$w = v_{\perp} \cdot t = 3.75 \times 30 = 112.5\mathrm{~m}$$
### Step 1: Final Conclusion
The width of the river is $112.5\mathrm{~m}$.
### Pattern Recognition
In river-boat crossing problems, river velocity ($v_r$) only causes drift along the bank; it has zero impact on crossing time or the crossing width calculations when the angle is defined relative to the flow. Use the vertical velocity component exclusively.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinematics
Q25 (2025)
A person travelling on a straight line moves with a uniform velocity $v_1$ for a distance $x$ and with a uniform velocity $v_2$ for the next $\frac{3}{2}x$ distance. The average velocity in this motion is $\frac{50}{7}\mathrm{~m/s}$. If $v_1$ is $5\mathrm{~m/s}$ then $v_2 = $ ____ $\mathrm{m/s}$.
### Related Formula
$$v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}$$
### Core Logic
Let's find the time taken for each section of the motion:
- First section of distance $x$ with velocity $v_1 = 5\mathrm{~m/s}$:
$$t_1 = \frac{x}{v_1} = \frac{x}{5}$$
- Second section of distance $\frac{3}{2}x$ with velocity $v_2$:
$$t_2 = \frac{3x/2}{v_2} = \frac{3x}{2v_2}$$
Total distance is:
$$d_{\text{total}} = x + \frac{3}{2}x = \frac{5}{2}x$$
Average velocity is:
$$v_{\text{avg}} = \frac{d_{\text{total}}}{t_1 + t_2} = \frac{\frac{5}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}}$$
We are given $v_{\text{avg}} = \frac{50}{7}\mathrm{~m/s}$. Equating the two values (noting $x$ cancels out):
$$\frac{50}{7} = \frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2v_2}}$$
Divide both sides by 5:
$$\frac{10}{7} = \frac{\frac{1}{2}}{\frac{1}{5} + \frac{3}{2v_2}} \implies 10 \left(\frac{1}{5} + \frac{3}{2v_2}\right) = \frac{7}{2}$$
$$2 + \frac{15}{v_2} = 3.5 \implies \frac{15}{v_2} = 1.5 \implies v_2 = \frac{15}{1.5} = 10\mathrm{~m/s}$$
### Step 1: Final Conclusion
The velocity $v_2$ is $10\mathrm{~m/s}$.
### Pattern Recognition
Never take simple arithmetic averages of velocities! Average velocity must always be calculated as $\frac{\text{Total Distance}}{\text{Total Time}}$. Because total distance and time intervals are proportional to $x$, $x$ cleanly cancels out.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinematics
Q11 (2025)
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 $\frac{nu^2}{25g}$ , where value of $n$ is:
(Given ' $g$' is the acceleration due to gravity).
- 6
- 18
- 12
- 24
### Related Formula
For a projectile with launch speed $u$ and angle $\theta$:
- Horizontal Range:
$$R = \frac{u^2 \sin(2\theta)}{g} = \frac{2 u^2 \sin\theta \cos\theta}{g}$$
- Maximum Height:
$$H = \frac{u^2 \sin^2\theta}{2g}$$
The general ratio linking range and maximum height is:
$$\tan\theta = \frac{4H}{R}$$
### Core Logic
Given state:
$$R = 3H \Rightarrow \frac{H}{R} = \frac{1}{3}$$
### Step 1: Determine the projection angle ($\theta$)
Substitute the ratio into the relation:
$$\tan\theta = 4 \left(\frac{H}{R}\right) = 4 \left(\frac{1}{3}\right) = \frac{4}{3}$$
This is a standard Pythagorean triangle angle:
$$\sin\theta = \frac{4}{5}, \quad \cos\theta = \frac{3}{5}$$
### Step 2: Compute the horizontal range ($R$)
$$R = \frac{2 u^2 \sin\theta \cos\theta}{g}$$
$$R = \frac{2 u^2 \left(\frac{4}{5}\right) \left(\frac{3}{5}\right)}{g} = \frac{24 u^2}{25 g}$$
Comparing this with the given format $\frac{nu^2}{25g}$:
$n = 24$
### Pattern Recognition
The relation $\tan\theta = 4H/R$ is an essential identity in projectile dynamics. Whenever $R = k H$, then $\tan\theta = 4/k$. Recognizing standard angles like $\tan\theta = 4/3$ or $3/4$ directly yields trigonometric values immediately.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Motion in a Plane
Q14 (2025)
A particle moves along the x-axis and has its displacement x varying with time t according to the equation $x=c_{0}(t^{2}-2)+c(t-2)^{2}$ where $c_{0}$ and c are constants of appropriate dimensions. Then, which of the following statements is correct?
- the acceleration of the particle is $2c_{0}$
- the acceleration of the particle is 2c
- the initial velocity of the particle is 4c
- the acceleration of the particle is $2(c+c_{0})$
### Related Formula
In rectilinear kinematics:
- Velocity:
$$v = \frac{dx}{dt}$$
- Acceleration:
$$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$$$
### Core Logic
Given position-time function:
$$x(t) = c_0 (t^2 - 2) + c (t - 2)^2$$
### Step 1: Differentiate once to get velocity ($v$)
$$v = \frac{dx}{dt} = \frac{d}{dt}\left[c_0(t^2 - 2)\right] + \frac{d}{dt}\left[c(t-2)^2\right]$$
$$v = c_0 (2t) + c \cdot 2(t-2) = 2 c_0 t + 2 c(t - 2)$$
### Step 2: Differentiate again to get acceleration ($a$)
$$a = \frac{dv}{dt} = \frac{d}{dt}\left[2 c_0 t + 2 c(t - 2)\right]$$
$$a = 2 c_0 + 2 c = 2(c + c_0)$$
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 $t$, the acceleration is constant and equal to $2 \times$ the coefficient of the $t^2$ term.
Rewriting $x(t)$:
$$x(t) = (c_0 + c)t^2 - 4ct + (4c - 2c_0)$$
The coefficient of $t^2$ is $(c_0 + c)$. Thus, acceleration is $2(c_0 + c)$ directly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Motion in a Straight Line
