Kinematics (Motion in 1D and 2D): JEE Main 2026 PYQs

Answer: Option (2)

Key idea: Acceleration from velocity field using convective derivative

1. Given velocity field: $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$ m/s
2. Use convective derivative: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{\partial \vec{v}}{\partial x}v_x + \frac{\partial \vec{v}}{\partial y}v_y + \frac{\partial \vec{v}}{\partial z}v_z$
3. Calculate: $\vec{a} = (-\hat{i})(-x) + (2\hat{j})(2y) + (-\hat{k})(-z) = x\hat{i} + 4y\hat{j} + z\hat{k}$
4. At point (1, 2, 4): $\vec{a} = \hat{i} + 8\hat{j} + 4\hat{k}$ m/s²
5. Magnitude: $|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{81} = 9$ m/s²

Answer: Option (2)

Key idea: Range of projectile motion and trigonometric ratios

1. Write the range formula: $R = \frac{u^2 \sin 2\theta}{g}$
2. For first projectile at $15°$: $R_1 = \frac{u^2 \sin 30°}{g} = \frac{u^2}{2g}$
3. For second projectile at $30°$: $R_2 = \frac{u^2 \sin 60°}{g} = \frac{u^2\sqrt{3}}{2g}$
4. Find the ratio: $\frac{R_1}{R_2} = \frac{u^2/(2g)}{u^2\sqrt{3}/(2g)} = \frac{1}{\sqrt{3}}$
5. Since $R_1:R_2 = 1:x$, we get $x = \sqrt{3}$

Answer: Option (3)

Key idea: Relative velocity in one dimension

1. Convert speeds to m/s: $v_A = 100$ km/h $= \frac{250}{9}$ m/s, $v_B = 80$ km/h $= \frac{200}{9}$ m/s
2. Stone speed in ground frame: $v_s = v_B + v$ (where $v$ is speed relative to car B)
3. Relative speed of stone w.r.t. car A: $v_{rel} = v_s - v_A = 5$ m/s
4. Substitute: $(v_B + v) - v_A = 5$
5. Solve: $v = v_A - v_B + 5 = \frac{250}{9} - \frac{200}{9} + 5 = \frac{95}{9}$ m/s
6. Convert to km/h: $v = \frac{95}{9} \times \frac{18}{5} = 38$ km/h

Answer: Option (2)

Key idea: Projectile Motion - Velocity and Angle from Position Functions

1. Given position functions: $x(t) = 24t$ and $y(t) = 43.6t - 4.9t^2$
2. Find velocity components by differentiation: $v_x = \frac{dx}{dt} = 24$ m/s and $v_y = \frac{dy}{dt} = 43.6 - 9.8t$ m/s
3. Calculate $v_y$ at $t = 2$ s: $v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24$ m/s
4. Find angle using $\tan\theta = \frac{v_y}{v_x} = \frac{24}{24} = 1$
5. Therefore, $\theta = \tan^{-1}(1) = 45°$

Answer: 8

Key idea: Maximum range of projectile motion occurs at 45° angle and equals u²/g

1. For projectile motion, range $R = \frac{u^2 \sin(2\theta)}{g}$
2. Maximum range occurs when $\sin(2\theta) = 1$, i.e., at $\theta = 45°$
3. Therefore, $R_{\text{max}} = \frac{u^2}{g}$
4. Substitute given values: $6.4 = \frac{u^2}{10}$
5. Solve for $u$: $u^2 = 64$, so $u = 8$ m/s

Answer: 13

Key idea: Kinematics - Free fall motion with condition v = g numerically

1. Given: Initial height $h_0 = 18$ m, initial velocity $u = 0$, $g = 10$ m/s²
2. Condition: Magnitude of velocity = Magnitude of acceleration, so $v = 10$ m/s
3. Let the ball fall through distance $s$ from top. Use $v^2 = u^2 + 2gs$
4. Substitute: $(10)^2 = 0 + 2(10)s \Rightarrow 100 = 20s \Rightarrow s = 5$ m
5. Height above ground $= 18 - 5 = 13$ m

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Key idea: Velocity-Time Graph Analysis: Distance vs Displacement

Answer: Option (2)

Key idea: Circular Motion with Energy Conservation

1. Step 1: At the highest point, the body exerts force $3mg$ on the plane, so normal force $N = 3mg$ acts on the body (toward center).
2. Step 2: Apply centripetal force equation at highest point: $N + mg = \frac{mv_{top}^2}{R}$, giving $4mg = \frac{mv_{top}^2}{R}$.
3. Step 3: Solve for velocity at top: $v_{top}^2 = 4gR$.
4. Step 4: Apply energy conservation from initial height $h$ to highest point (at height $2R$): $mgh = \frac{1}{2}mv_{top}^2 + mg(2R)$.
5. Step 5: Substitute $v_{top}^2 = 4gR$: $mgh = 2mgR + 2mgR = 4mgR$.
6. Step 6: Therefore $h = 4R$, so $\alpha = 4$.

Answer: tan⁻¹(9/8) ≈ 48.4°

Key idea: Circular motion in non-inertial reference frame; pseudo forces on pendulum

1. Convert speed: $v = 54$ km/h $= 54 \times \frac{5}{18} = 15$ m/s
2. In the car's frame, the bob experiences centrifugal force $\frac{mv^2}{r}$ (horizontal) and weight $mg$ (vertical)
3. At equilibrium, resolve forces: $T\sin\theta = \frac{mv^2}{r}$ and $T\cos\theta = mg$
4. Divide the equations: $\tan\theta = \frac{v^2}{rg}$
5. Substitute values: $\tan\theta = \frac{(15)^2}{20 \times 10} = \frac{225}{200} = \frac{9}{8}$
6. Calculate angle: $\theta = \tan^{-1}\left(\frac{9}{8}\right) \approx 48.4°$

Answer: Option (4)

Key idea: Kinematics: Motion under gravity with initial velocity; relative motion

1. Stone is dropped from balloon at height $h_0 = 75$ m with initial upward velocity $v_0 = 10$ m/s (same as balloon)
2. Apply equation of motion for stone: $h = h_0 + v_0t - \frac{1}{2}gt^2$. When stone hits ground, $h = 0$
3. Substitute values: $0 = 75 + 10t - 5t^2$, which gives $t^2 - 2t - 15 = 0$
4. Solve quadratic equation: $t = \frac{2 + \sqrt{4 + 60}}{2} = \frac{2 + 8}{2} = 5$ seconds
5. Balloon continues upward with constant velocity: Distance traveled $= 10 \times 5 = 50$ m
6. Final height of balloon $= 75 + 50 = 125$ m

Answer: Option (1)

Key idea: Projectile Motion - Complementary Angles with Same Range

1. For projectile motion, time of flight $T = \frac{2u \sin \theta}{g}$. From $t_1 = 5$ s: $u \sin \theta_1 = 25$ m/s
2. From $t_2 = 10$ s: $u \sin \theta_2 = 50$ m/s
3. For same range with same speed, angles are complementary: $\theta_2 = 90° - \theta_1$, so $\sin \theta_2 = \cos \theta_1$
4. Using $\sin^2 \theta_1 + \cos^2 \theta_1 = 1$: $(u \sin \theta_1)^2 + (u \cos \theta_1)^2 = u^2$, giving $25^2 + 50^2 = u^2 = 3125$
5. Calculate $\sin 2\theta_1 = 2 \sin \theta_1 \cos \theta_1 = 2 \times \frac{25}{u} \times \frac{50}{u} = \frac{2500}{3125} = \frac{4}{5}$
6. Range $R = \frac{u^2 \sin 2\theta_1}{g} = \frac{3125 \times 4/5}{10} = \frac{2500}{10} = 250$ m