Grade 11 : Physics - Motion (Straight/Plane) (Set 2)

Explanation:Displacement is the shortest straight-line distance between the initial and final positions. For half a circle of radius $R$, the straight line distance is equal to the diameter: $\text{Displacement} = 2R = 2 \times 7m = 14m$.

Explanation:When equal distances are covered with speeds $v_1$ and $v_2$, average speed is given by harmonic mean: $v_{\text{avg}} = \frac{2 v_1 v_2}{v_1 + v_2} = \frac{2 \times 30 \times 60}{30 + 60} = \frac{3600}{90} = 40km/h$.

Explanation:Using first kinematic equation: $v = u + at$. Given $u = 0m/s$, $a = 2m/s^2$, $t = 5\text{ s}$. Thus, $v = 0 + (2 \times 5) = 10m/s$.

Explanation:At maximum height, final velocity $v = 0$. Using $v^2 = u^2 - 2gh$: $0 = 20^2 - 2(10)h \implies 20h = 400 \implies h = 20m$.

Explanation:Distance in the $n^{\text{th}}$ second is $S_n = u + \frac{a}{2}(2n - 1)$. Here $u = 0$, $a = 4$, $n = 3$. $S_3 = 0 + \frac{4}{2}(2 \times 3 - 1) = 2(5) = 10m$.

Explanation:Displacement is equal to the area under the velocity-time graph. $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6\text{ s} \times 10m/s = 30m$.

Explanation:The slope of a position-time graph is $\frac{\Delta x}{\Delta t}$, which defines velocity.

Explanation:Horizontal range is $R = \frac{u^2 \sin(2\theta)}{g}$. Range is maximum when $\sin(2\theta) = 1 \implies 2\theta = 90^\circ \implies \theta = 45^\circ$.

Explanation:$H_{\text{max}} = \frac{u^2 \sin^2\theta}{2g} = \frac{30^2 \times (\sin 30^\circ)^2}{2 \times 10} = \frac{900 \times 0.25}{20} = 11.25m$.

Explanation:Time of flight is $T = \frac{2 u \sin\theta}{g} = \frac{2 \times 20 \times \sin 30^\circ}{10} = \frac{40 \times 0.5}{10} = 2\text{ s}$.

Explanation:When two objects move in opposite directions, relative speed is the sum of their individual speeds: $v_{AB} = v_A - (-v_B) = v_A + v_B = 50 + 40 = 90km/h$.

Explanation:Centripetal acceleration is given by $a_c = \frac{v^2}{R}$. Substituting $v = 8m/s$ and $R = 4m$: $a_c = \frac{8^2}{4} = \frac{64}{4} = 16m/s^2$.

Explanation:Acceleration is defined as the rate of change of velocity ($a = \frac{\Delta v}{\Delta t}$). Constant velocity implies $\Delta v = 0$, so acceleration is zero.

Explanation:Using $v^2 = u^2 - 2as$ with final velocity $v = 0$: $0 = 10^2 - 2a(20) \implies 40a = 100 \implies a = 2.5m/s^2$.

Explanation:Speed is the magnitude of the velocity vector: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10m/s$.

Explanation:For free fall from rest ($u = 0$), $h = \frac{1}{2}gt^2 \implies 45 = 5t^2 \implies t^2 = 9 \implies t = 3\text{ s}$.

Explanation:Angles $\theta_1 = 30^\circ$ and $\theta_2 = 60^\circ$ are complementary ($\theta_1 + \theta_2 = 90^\circ$). For complementary angles at equal projection speed, horizontal ranges are equal, so the ratio is $1 : 1$.

Explanation:In uniform circular motion, velocity is tangential to the circle, while centripetal acceleration is directed radially inward. Hence, they are perpendicular ($90^\circ$).

Explanation:By definition, $a = \frac{dv}{dt} \implies \int a \, dt = \Delta v$. Thus, the area under the acceleration-time graph represents the change in velocity.

Explanation:The body covers equal distances (10 m) in equal intervals of time (2 s), which means speed is constant (5 m/s). This defines uniform motion.