Grade 11 : Physics - Motion (Straight/Plane) (Set 1)

Explanation: Displacement has both magnitude and direction, making it a vector quantity. Distance, speed, and time are scalar quantities possessing only magnitude.

Explanation: Time taken for first half $t_1 = \frac{60}{30} = 2\text{ h}$. Time taken for second half $t_2 = \frac{60}{60} = 1\text{ h}$. Total time $t = 3\text{ h}$. Total distance $d = 120km$. Average speed $v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{120}{3} = 40km/h$.

Explanation: Using the first equation of motion $v = u + at$: Here $u = 0m/s$, $a = 2m/s^2$, and $t = 5\text{ s}$. Therefore, $v = 0 + (2)(5) = 10m/s$.

Explanation: Using the displacement equation $h = ut + \frac{1}{2}gt^2$: Since $u = 0$, $20 = \frac{1}{2}(10)t^2 \implies 20 = 5t^2 \implies t^2 = 4 \implies t = 2\text{ s}$.

Explanation: By definition, the rate of change of position with respect to time $\frac{dx}{dt}$ is velocity. Hence, the slope of a position-time graph yields velocity.

Explanation: The integral of velocity with respect to time $\int v \, dt$ equals displacement. Therefore, the area under the $v-t$ curve gives the net displacement.

Explanation: Using $v^2 = u^2 - 2gH$: At maximum height, $v = 0$. So $0 = u^2 - 2gH \implies H = \frac{u^2}{2g}$.

Explanation: Velocity $v(t) = \frac{dx}{dt} = 8t - 2$. Acceleration $a(t) = \frac{dv}{dt} = 8m/s^2$, which is constant.

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

Explanation: At the highest point, the vertical component of velocity becomes zero ($v_y = 0$). Only the horizontal component remains constant ($v_x = u\cos\theta$).

Explanation: $\vec{v}(t) = \frac{d\vec{r}}{dt} = 3\hat{i} + 4t\hat{j}$. At $t = 1\text{ s}$, $\vec{v} = 3\hat{i} + 4\hat{j}$. Magnitude $|\vec{v}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5m/s$.

Explanation: Relative speed $v_{\text{rel}} = 15 + 25 = 40m/s$. Total distance to cover $d = 100 + 100 = 200m$. Time $t = \frac{d}{v_{\text{rel}}} = \frac{200}{40} = 5\text{ s}$.

Explanation: Instantaneous velocity $v = \frac{ds}{dt} = 6t$. Evaluating at $t = 3\text{ s}$: $v = 6(3) = 18m/s$.

Explanation: Gravity acts vertically downwards. With no horizontal force or air resistance acting on the projectile, the horizontal acceleration is identically zero.

Explanation: Distance in $n\text{-th}$ second: $s_n = u + \frac{a}{2}(2n - 1)$. With $u = 0$, $a = 3$, $n = 4$: $s_4 = 0 + \frac{3}{2}(2(4) - 1) = 1.5 \times 7 = 10.5m$.

Explanation: Displacement is the shortest straight-line distance between initial and final points, so $|\text{Displacement}| \le \text{Distance}$. Therefore, $\frac{|\text{Displacement}|}{\text{Distance}} \le 1$.

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

Explanation: Acceleration is $a = \frac{dv}{dt}$. Integrating gives $\int a \, dt = \Delta v$. Hence, the area under the $a-t$ graph represents the change in velocity.

Explanation: When moving in the same direction, relative velocity $v_{PQ} = v_P - v_Q = 70km/h - 50km/h = 20km/h$.

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