Grade 11 : Physics - Motion (Straight/Plane) (Set 4)

Explanation: Average velocity $v_{avg}$ is given by $\frac{x(t_2) - x(t_1)}{t_2 - t_1}$. Here $x(6) = 44m$ and $x(2) = 12m$. Thus, $v_{avg} = \frac{44 - 12}{6 - 2} = \frac{32}{4} = 8m/s$.

Explanation: Angular velocity $\omega = \frac{v}{r}$. For Runner A: $\omega_A = \frac{5}{100} = 0.05\text{ rad/s}$. For Runner B: $\omega_B = \frac{6}{150} = 0.04\text{ rad/s}$. The ratio $\frac{\omega_A}{\omega_B} = \frac{0.05}{0.04} = 1.25$.

Explanation: At maximum height, the vertical velocity component is zero, so the speed equals the horizontal component: $v = u \cos\theta$. Given $v = \frac{u}{2}$, we have $u \cos\theta = \frac{u}{2} \Rightarrow \cos\theta = 0.5$, which yields $\theta = 60^\circ$.

Explanation: The acceleration is constant at $a = \frac{4 - 0}{1} = 4m/s^2$. Displacement $s = u t + \frac{1}{2} a t^2 = 0(4) + \frac{1}{2}(4)(4^2) = 32m$. Alternatively, area under $v$-$t$ graph $= \frac{1}{2}(0 + 16) \times 4 = 32m$.

Explanation: To travel straight across, the boat's velocity component along the river must cancel the current: $v_x = v_r = 3m/s$. The resultant velocity across the river is $v_y = \sqrt{v_b^2 - v_r^2} = \sqrt{5^2 - 3^2} = 4m/s$. Time taken $t = \frac{W}{v_y} = \frac{400}{4} = 100\text{ s}$.

Explanation: Velocity $v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12$. Acceleration $a(t) = \frac{dv}{dt} = 12t - 18$. Setting $a(t) = 0 \Rightarrow 12t - 18 = 0 \Rightarrow t = 1.5\text{ s}$.

Explanation: Total height $H = \frac{1}{2}gT^2$. Distance in first $(T-1)$ seconds is $H' = \frac{1}{2}g(T-1)^2 = H - \frac{7}{16}H = \frac{9}{16}H$. Therefore, $\frac{\frac{1}{2}g(T-1)^2}{\frac{1}{2}gT^2} = \frac{9}{16} \Rightarrow \frac{T-1}{T} = \frac{3}{4} \Rightarrow 4T - 4 = 3T \Rightarrow T = 4\text{ s}$.

Explanation: Tangential acceleration $a_t = \frac{dv}{dt} = 2m/s^2$. At $t = 2\text{ s}$, speed $v = 4m/s$. Centripetal acceleration $a_c = \frac{v^2}{R} = \frac{4^2}{4} = 4m/s^2$. Total acceleration $a = \sqrt{a_t^2 + a_c^2} = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}m/s^2$.

Explanation: $\vec{v}(t) = \frac{d\vec{r}}{dt} = (6t - 4)\hat{i} + 2\hat{j}$. At $t = 2\text{ s}$, $\vec{v}(2) = (12 - 4)\hat{i} + 2\hat{j} = 8\hat{i} + 2\hat{j}$. Magnitude $|v| = \sqrt{8^2 + 2^2} = \sqrt{68} = 2\sqrt{17}m/s$.

Explanation: $h_1 = \frac{u^2 \sin^2\theta_1}{2g} \Rightarrow 5 = \frac{400 \sin^2\theta_1}{20} \Rightarrow \sin^2\theta_1 = \frac{1}{4} \Rightarrow \theta_1 = 30^\circ$. Complementary angle $\theta_2 = 60^\circ$. $h_2 = \frac{u^2 \sin^2 60^\circ}{2g} = \frac{400 \times (3/4)}{20} = 15m$.

Explanation: Let East be $\hat{i}$ and North be $\hat{j}$. Velocity vectors: $\vec{v}_A = 30\hat{i}$, $\vec{v}_B = 40\hat{j}$. Relative velocity $\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A = 40\hat{j} - 30\hat{i}$. Magnitude $|\vec{v}_{B/A}| = \sqrt{40^2 + (-30)^2} = 50km/h$.

Explanation: Equation: $x(t) = ut + \frac{1}{2}at^2$. For $t=1$: $5 = u + 0.5a$. For $t=2$: $16 = 2u + 2a \Rightarrow 8 = u + a$. Subtracting equations gives $0.5a = 3 \Rightarrow a = 6m/s^2$. Then $u = 8 - 6 = 2m/s$.

Explanation: Vertical motion: $h = \frac{1}{2}gt^2 \Rightarrow 45 = 5t^2 \Rightarrow t = 3\text{ s}$. Horizontal motion is uniform: $x = v_0 t = 15 \times 3 = 45m$.

Explanation: Distance along half circle $= \pi r = \frac{22}{7} \times 7 = 22m$. Magnitude of displacement $= 2r = 14m$. Ratio $= \frac{22}{14} = \frac{11}{7}$.

Explanation: Relative initial speed $u_{rel} = 20 - 10 = 10m/s$. Relative deceleration $a_{rel} = -1m/s^2$. Distance covered to match speeds ($v_{rel} = 0$): $s_{rel} = \frac{u_{rel}^2}{-2a_{rel}} = \frac{100}{2} = 50m$. Since $50m < 100m$, Car A never reaches Car B. Minimum separation is $100 - 50 = 50m$.

Explanation: Horizontal velocity remains constant: $v_x = 30m/s$. Vertical velocity $v_y = 40 - gt = 40 - 10(3) = 10m/s$. Speed $v = \sqrt{v_x^2 + v_y^2} = \sqrt{30^2 + 10^2} = \sqrt{1000} = 10\sqrt{10}m/s$.

Explanation: Velocity of rain relative to person $\vec{v}_{r/p} = \vec{v}_r - \vec{v}_p = -12\hat{j} - 5\hat{i}$. Angle $\theta$ with vertical is $\tan\theta = \frac{|v_{p}|}{|v_{r}|} = \frac{5}{12} \Rightarrow \theta = \tan^{-1}(5/12)$.

Explanation: $S_1 = \frac{1}{2} a (2^2) = 2a$. Total distance in 4 s is $S_{total} = \frac{1}{2} a (4^2) = 8a$. Distance in second 2 s interval $S_2 = S_{total} - S_1 = 8a - 2a = 6a$. Thus ratio $S_1 : S_2 = 2a : 6a = 1 : 3$.

Explanation: Speed at time $t$ is $v = a_t t = 3t$. Radial acceleration $a_r = \frac{v^2}{R} = \frac{(3t)^2}{50} = \frac{9t^2}{50}$. Setting $a_r = a_t \Rightarrow \frac{9t^2}{50} = 3 \Rightarrow t^2 = \frac{150}{9} = \frac{50}{3} \approx 16.67 \Rightarrow t \approx 4.08\text{ s}$.

Explanation: Time for first leg $t_1 = \frac{600}{6} = 100\text{ s}$. Time for second leg $t_2 = \frac{200}{4} = 50\text{ s}$. Total time $T = 150\text{ s}$. Net displacement $\Delta x = 600 - 200 = 400m$. Average velocity magnitude $= \frac{400}{150} = \frac{8}{3} \approx 2.67m/s$.