Two cars P and Q are moving on a road in the same direction. Acceleration of car P increases linearly with time whereas car Q moves with a glass constant acceleration. Both cars cross each other at time t = 0$t = 0$ , for the first time. The maximum possible number of crossing(s) (including the crossing at t = 0$t = 0$ ) is ______.

### Related Formula Maximum height of a projectile: H_textmax = fracu^2 sin^2 theta2g$$H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}$$ where theta$\theta$ is the angle of projection measured from the horizontal ground. ### Core Logic The angle of projection in this question is measured from the **vertical axis** as phi$\phi$. This relates to the horizontal projection angle theta$\theta$ as: theta = 90^circ - phi$$\theta = 90^{\circ} - \phi$$ Substitute this into the maximum height formula: h_m = fracu^2 sin^2 (90^circ - phi)2g = fracu^2 cos^2 phi2g$$h_m = \frac{u^2 \sin^2 (90^{\circ} - \phi)}{2g} = \frac{u^2 \cos^2 \phi}{2g}$$ h_m(phi) propto cos^2 phi$$h_m(\phi) \propto \cos^2 \phi$$ ### Step 1: Analyzing Boundary Conditions Let's check the boundary values of h_m$h_m$ for phi in [0, 90^circ]$\phi \in [0, 90^{\circ}]$: - At phi = 0^circ$\phi = 0^{\circ}$ (fired straight up vertically): h_m(0) = fracu^22g = textMaximum height$$h_m(0) = \frac{u^2}{2g} = \text{Maximum height}$$ - At phi = 90^circ$\phi = 90^{\circ}$ (fired horizontally along the ground): h_m(90^circ) = 0$$h_m(90^{\circ}) = 0$$ Let's evaluate the behavior of h_m(phi)$h_m(\phi)$: - At small angles phi approx 0$\phi \approx 0$, the slope fracd h_mdphi = -fracu^22g sin(2phi) approx 0$\frac{d h_m}{d\phi} = -\frac{u^2}{2g} \sin(2\phi) \approx 0$. This means the curve starts with a horizontal slope (flat top). - As phi$\phi$ increases towards 90^circ$90^{\circ}$, h_m$h_m$ falls smoothly, reaching 0 at 90^circ$90^{\circ}$ with flat slope again. This sinusoidal falloff perfectly matches **Curve (3)**.

### Pattern Recognition Always read the coordinate origin carefully! Measuring angle from the vertical instead of the horizontal turns sin^2theta$\sin^2\theta$ into cos^2phi$\cos^2\phi$. Checking extremes: vertical shot (0^circ$0^{\circ}$) yields maximum height, while flat shot (90^circ$90^{\circ}$) yields zero height. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane: Projectile Motion

### Related Formula vecv = fracdvecSdt$$\vec{v} = \frac{d\vec{S}}{dt}$$ ### Core Logic Differentiating the position vector with respect to time gives the velocity vector: vecv = fracddt(2t^2hatj + 5hatk) = 4thatj$$\vec{v} = \frac{d}{dt}(2t^{2}\hat{j} + 5\hat{k}) = 4t\hat{j}$$ ### Step 1: Evaluation at given time Substitute t = 1text s$t = 1\text{ s}$ into the velocity expression: vecv = 4(1)hatj = 4hatjtext m/s$$\vec{v} = 4(1)\hat{j} = 4\hat{j}\text{ m/s}$$ ### Pattern Recognition Constant term vectors (like 5hatk$5\hat{k}$) drop out during differentiation. The resulting velocity only has a component along the hatj$\hat{j}$ direction, confirming motion purely parallel to the y-axis at that instant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula v_x^2 - u_x^2 = 2 a_x x$$v_x^2 - u_x^2 = 2 a_x x$$ v_y = u_y + a_y t$$v_y = u_y + a_y t$$ ### Core Logic Analyze the motion along the x-axis first to find the final x-velocity component (u_x = 5text m/s$u_x = 5\text{ m/s}$, a_x = 3text m/s^2$a_x = 3\text{ m/s}^2$, x = 84text m$x = 84\text{ m}$): v_x^2 - 5^2 = 2(3)(84)$$v_x^2 - 5^2 = 2(3)(84)$$ v_x^2 - 25 = 504 implies v_x^2 = 529 implies v_x = 23text m/s$$v_x^2 - 25 = 504 \implies v_x^2 = 529 \implies v_x = 23\text{ m/s}$$ ### Step 1: Compute time interval Using the velocity relation along the x-axis: v_x = u_x + a_x t implies 23 = 5 + 3t implies 3t = 18 implies t = 6text s$$v_x = u_x + a_x t \implies 23 = 5 + 3t \implies 3t = 18 \implies t = 6\text{ s}$$ ### Step 2: Evaluate y-axis velocity component Since u_y = 0$u_y = 0$ and a_y = 2text m/s^2$a_y = 2\text{ m/s}^2$, calculate v_y$v_y$ at t = 6text s$t = 6\text{ s}$: v_y = 0 + 2(6) = 12text m/s$$v_y = 0 + 2(6) = 12\text{ m/s}$$ ### Step 3: Total net speed calculation v^2 = v_x^2 + v_y^2 = 23^2 + 12^2 = 529 + 144 = 673$$v^2 = v_x^2 + v_y^2 = 23^2 + 12^2 = 529 + 144 = 673$$ v = sqrt673text m/s$$v = \sqrt{673}\text{ m/s}$$ Comparing this with sqrta$\sqrt{a}$ yields a = 673$a = 673$. ### Pattern Recognition Splitting coordinates explicitly isolates vector calculation lines cleanly, optimizing equation selections relative to independent components. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula v = fracdxdt$$v = \frac{dx}{dt}$$ a = fracdvdt = vfracdvdx$$a = \frac{dv}{dt} = v\frac{dv}{dx}$$ ### Step 1: Differentiate with respect to time Given the equation: t = alpha x^2 + beta x$$t = \alpha x^2 + \beta x$$ Differentiating with respect to time t$t$: fracdtdt = fracddt(alpha x^2 + beta x)$$\frac{dt}{dt} = \frac{d}{dt}(\alpha x^2 + \beta x)$$ 1 = 2alpha x fracdxdt + beta fracdxdt$$1 = 2\alpha x \frac{dx}{dt} + \beta \frac{dx}{dt}$$ 1 = (2alpha x + beta) v$$1 = (2\alpha x + \beta) v$$ v = (2alpha x + beta)^-1$$v = (2\alpha x + \beta)^{-1}$$ ### Step 2: Calculate Acceleration Now, acceleration a = fracdvdt$a = \frac{dv}{dt}$. Differentiating v$v$ with respect to time t$t$: a = fracddt left[ (2alpha x + beta)^-1 right]$$a = \frac{d}{dt} \left[ (2\alpha x + \beta)^{-1} \right]$$ a = -1(2alpha x + beta)^-2 cdot fracddt(2alpha x + beta)$$a = -1(2\alpha x + \beta)^{-2} \cdot \frac{d}{dt}(2\alpha x + \beta)$$ a = -(2alpha x + beta)^-2 cdot (2alpha) fracdxdt$$a = -(2\alpha x + \beta)^{-2} \cdot (2\alpha) \frac{dx}{dt}$$ Substitute v$v$ and (2alpha x + beta)^-2 = v^2$(2\alpha x + \beta)^{-2} = v^2$: a = -(v^2) cdot (2alpha) cdot v$$a = -(v^2) \cdot (2\alpha) \cdot v$$ a = -2alpha v^3$$a = -2\alpha v^3$$ ### Pattern Recognition Standard kinematic shortcut: Whenever t = Ax^2 + Bx$t = Ax^2 + Bx$, v = (2Ax+B)^-1$v = (2Ax+B)^{-1}$ and a = -2A v^3$a = -2A v^3$. Memorizing this directly saves derivation time during the exam. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula t = sqrtfrac2dg$$t = \sqrt{\frac{2d}{g}}$$ ### Core Logic

The body falls freely from height H$H$ to height h$h$. The distance fallen is (H - h)$(H - h)$. Time taken to fall this distance: t_1 = sqrtfrac2(H - h)g$$t_1 = \sqrt{\frac{2(H - h)}{g}}$$ After elastic impact, the vertical velocity becomes zero (the entire velocity is directed horizontally). From height h$h$, it now acts as a horizontal projectile. The time taken to reach the ground vertically from height h$h$ is: t_2 = sqrtfrac2hg$$t_2 = \sqrt{\frac{2h}{g}}$$ Total time of flight T = t_1 + t_2$T = t_1 + t_2$: T = sqrtfrac2(H - h)g + sqrtfrac2hg$$T = \sqrt{\frac{2(H - h)}{g}} + \sqrt{\frac{2h}{g}}$$ ### Step 2: Maximizing Time To find the maximum time T$T$, differentiate T$T$ with respect to h$h$ and equate to zero: fracdTdh = sqrtfrac2g left( frac-12sqrtH - h + frac12sqrth right) = 0$$\frac{dT}{dh} = \sqrt{\frac{2}{g}} \left( \frac{-1}{2\sqrt{H - h}} + \frac{1}{2\sqrt{h}} \right) = 0$$ frac12sqrth = frac12sqrtH - h$$\frac{1}{2\sqrt{h}} = \frac{1}{2\sqrt{H - h}}$$ sqrtH - h = sqrth$$\sqrt{H - h} = \sqrt{h}$$ Squaring both sides: H - h = h$H - h = h$ H = 2h$H = 2h$ fracHh = 2$$\frac{H}{h} = 2$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics