### Related Formula v_perp = v_br sintheta$v_{\perp} = v_{br} \sin\theta$ w = v_perp cdot t$w = v_{\perp} \cdot t$ ### Core Logic Let the West-to-East river flow direction be along the positive x$x$-axis. The boat's velocity relative to water is v_br = 27mathrm~km/h$v_{br} = 27\mathrm{~km/h}$ at an angle of 150^circ$150^{circ}$ with respect to the river flow (which is 30^circ$30^{circ}$ 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$y$-axis): v_perp = v_br sin(150^circ) = 27 sin(150^circ) = 27 times frac12 = 13.5mathrm~km/h$v_{\perp} = v_{br} \sin(150^{circ}) = 27 \sin(150^{circ}) = 27 \times \frac{1}{2} = 13.5\mathrm{~km/h}$ Convert this velocity to SI units: v_perp = 13.5 times frac518mathrm~m/s = 3.75mathrm~m/s$v_{\perp} = 13.5 \times \frac{5}{18}\mathrm{~m/s} = 3.75\mathrm{~m/s}$ Given the crossing time is half a minute (t = 30mathrm~s$t = 30\mathrm{~s}$): w = v_perp cdot t = 3.75 times 30 = 112.5mathrm~m$w = v_{\perp} \cdot t = 3.75 \times 30 = 112.5\mathrm{~m}$ ### Step 1: Final Conclusion The width of the river is $ ### Step 1: Final Conclusion The width of the river is $112.5\mathrm{~m}. ### Pattern Recognition In river-boat crossing problems, river velocity ($. ### 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 null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula v_textavg = fractextTotal DistancetextTotal Time$v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}$ ### Core Logic Let's find the time taken for each section of the motion: - First section of distance x$x$ with velocity v_1 = 5mathrm~m/s$v_1 = 5\mathrm{~m/s}$: t_1 = fracxv_1 = fracx5$t_1 = \frac{x}{v_1} = \frac{x}{5}$ - Second section of distance frac32x$\frac{3}{2}x$ with velocity v_2$v_2$: t_2 = frac3x/2v_2 = frac3x2v_2$t_2 = \frac{3x/2}{v_2} = \frac{3x}{2v_2}$ Total distance is: d_texttotal = x + frac32x = frac52x$d_{\text{total}} = x + \frac{3}{2}x = \frac{5}{2}x$ Average velocity is: v_textavg = fracd_texttotalt_1 + t_2 = fracfrac52xfracx5 + frac3x2v_2$v_{\text{avg}} = \frac{d_{\text{total}}}{t_1 + t_2} = \frac{\frac{5}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}}$ We are given v_textavg = frac507mathrm~m/s$v_{\text{avg}} = \frac{50}{7}\mathrm{~m/s}$. Equating the two values (noting x$x$ cancels out): frac507 = fracfrac52frac15 + frac32v_2$\frac{50}{7} = \frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2v_2}}$ Divide both sides by 5: frac107 = fracfrac12frac15 + frac32v_2 implies 10 left(frac15 + frac32v_2right) = frac72$\frac{10}{7} = \frac{\frac{1}{2}}{\frac{1}{5} + \frac{3}{2v_2}} \implies 10 \left(\frac{1}{5} + \frac{3}{2v_2}\right) = \frac{7}{2}$ 2 + frac15v_2 = 3.5 implies frac15v_2 = 1.5 implies v_2 = frac151.5 = 10mathrm~m/s$2 + \frac{15}{v_2} = 3.5 \implies \frac{15}{v_2} = 1.5 \implies v_2 = \frac{15}{1.5} = 10\mathrm{~m/s}$ ### Step 1: Final Conclusion The velocity v_2$v_2$ is 10mathrm~m/s$10\mathrm{~m/s}$. ### Pattern Recognition Never take simple arithmetic averages of velocities! Average velocity must always be calculated as fractextTotal DistancetextTotal Time$\frac{\text{Total Distance}}{\text{Total Time}}$. Because total distance and time intervals are proportional to x$x$, x$x$ cleanly cancels out. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula H = fracu^2 sin^2theta2g quad textand quad T = frac2u sinthetag$H = \frac{u^2 \sin^2\theta}{2g} \quad \text{and} \quad T = \frac{2u \sin\theta}{g}$ where, H$H$ = maximum height reached T$T$ = total time of flight u$u$ = initial projection velocity theta$\theta$ = angle of projection ### Core Logic From the formulas, we see that: H propto sin^2theta quad textand quad T propto sintheta$H \propto \sin^2\theta \quad \text{and} \quad T \propto \sin\theta$ Thus, we can directly link time of flight to the square root of the maximum height: T propto sqrtH implies fracT_1T_2 = sqrtfracH_1H_2$T \propto \sqrt{H} \implies \frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}}$ ### Step 1: Compute Ratio Given: H_1 = 8 H_2 implies fracH_1H_2 = 8$H_1 = 8 H_2 \implies \frac{H_1}{H_2} = 8$ Substitute this ratio: fracT_1T_2 = sqrt8 = 2sqrt2$\frac{T_1}{T_2} = \sqrt{8} = 2\sqrt{2}$ Thus, the ratio is 2sqrt2 : 1$2\sqrt{2} : 1$. ### Pattern Recognition Sees: Projectile heights ratio → Time of flight ratio. Shortcut: Since H propto u_y^2$H \propto u_y^2$ and T propto u_y$T \propto u_y$, we have T propto sqrtH$T \propto \sqrt{H}$. If the height is 8 times larger, the flying time is sqrt8 = 2sqrt2$\sqrt{8} = 2\sqrt{2}$ times larger. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula a_P = kt implies v_P = frac12kt^2 + v_0P implies x_P = frac16kt^3 + v_0Pt + x_0$a_P = kt \implies v_P = \frac{1}{2}kt^2 + v_{0P} \implies x_P = \frac{1}{6}kt^3 + v_{0P}t + x_0$ a_Q = a implies v_Q = at + v_0Q implies x_Q = frac12at^2 + v_0Qt + x_0$a_Q = a \implies v_Q = at + v_{0Q} \implies x_Q = \frac{1}{2}at^2 + v_{0Q}t + x_0$ ### Core Logic To find the relative crossings, set the position functions equal: x_P(t) = x_Q(t)$x_P(t) = x_Q(t)$. frac16kt^3 + v_0Pt = frac12at^2 + v_0Qt$\frac{1}{6}kt^3 + v_{0P}t = \frac{1}{2}at^2 + v_{0Q}t$ frac16kt^3 - frac12at^2 + (v_0P - v_0Q)t = 0$\frac{1}{6}kt^3 - \frac{1}{2}at^2 + (v_{0P} - v_{0Q})t = 0$ This is a cubic polynomial equation in terms of time t$t$. A cubic function can yield a maximum of 3 distinct real roots.

Depending on the initial velocity profiles (v_0P, v_0Q$v_{0P}, v_{0Q}$), the curves can intersect at t=0$t=0$ and potentially create up to two additional crossing paths downstream as acceleration profiles cross over. Thus, the maximum possible number of total crossings is exactly 3. ### Pattern Recognition Crossings correspond mathematically to intersections of relative displacement polynomials. A linear acceleration profile produces a cubic position curve (t^3$t^3$), which can cross a quadratic curve (t^2$t^2$) at up to 3 distinct coordinate points. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula For realistic physical motion in one dimension: 1. Time t$t$ can never be negative during a normal positive time sequence, and cannot flow backwards. 2. Total distance covered can never decrease over time. 3. A particle cannot have two different values of position or velocity at the exact same instant of time. ### Core Logic Let us analyze each curve: - **Curve (A)** (Phase phi$\phi$ vs Time t$t$): Represents phi = kt + C$\phi = kt + C$, which is a valid linear relationship of phase over time (e.g., in Simple Harmonic Motion x = Asin(kt + C)$x = A\sin(kt + C)$). (Valid) - **Curve (B)** (Velocity v$v$ vs Displacement x$x$): A closed loop, which represents symmetric harmonic-type oscillation. For example, v^2 + omega^2 x^2 = textconst$v^2 + \omega^2 x^2 = \text{const}$ (ellipse) is a perfectly physically valid 1D SHM velocity-displacement phase portrait. (Valid) - **Curve (C)** (Velocity vs Time): The curve enters into the negative time quadrant. Time cannot go backwards or exist in negative values relative to starting sequence in standard physical scenarios. (Invalid) - **Curve (D)** (Total Distance vs Time): Represents total distance increasing over time. Total distance is a non-decreasing function of time (d(d)/dt ge 0$d(d)/dt \ge 0$). Thus, this curve is physically valid. (Valid) ### Step 1: Conclusion Therefore, curves A, B, and D possibly represent physical one-dimensional motion. The correct option is (1). ### Pattern Recognition Quick check for graph validity: 1. Time cannot run backwards (ruling out C). 2. Total distance can never decrease (D is valid because it strictly goes upwards). 3. v$v$ vs x$x$ can be circular/elliptical in SHM (B is valid). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line