### Related Formula 1. Distance: Total actual path length covered. 2. Displacement: Shortest straight-line distance connecting the initial and final position. 3. Circumference of a complete circle = 2pi r$2\pi r$ 4. Semicircular arc length = pi r$\pi r$ ### Core Logic The trajectory is defined by the sequence of points A to B to A to B$A \to B \to A \to B$: 1. **Distance Travelled:** - Segment 1 (A to B$A \to B$): Semicircular path of length pi r$\pi r$ - Segment 2 (B to A$B \to A$): Semicircular path of length pi r$\pi r$ - Segment 3 (A to B$A \to B$): Semicircular path of length pi r$\pi r$ - Total distance: textDistance = pi r + pi r + pi r = 3pi r$\text{Distance} = \pi r + \pi r + \pi r = 3\pi r$ 2. **Displacement:** - Initial position: A$A$ - Final position: B$B$ - Since points A$A$ and B$B$ represent diametrically opposite positions on the circle, the shortest distance between them is equal to the diameter of the circle: textDisplacement = 2r$\text{Displacement} = 2r$ ### Step 1: Write result The actual distance travelled is 3pi r$3\pi r$ and the magnitude of displacement is 2r$2r$. ### Pattern Recognition Sees: Circular kinematics path tracing. Trap: Accidentally substituting straight chords for the distance arcs, or assuming the loop returns fully to A (which would yield a zero displacement). Shortcut: The path ends at B. Since 3 half-loops are made, distance = 3 times pi r = 3pi r$3 \times \pi r = 3\pi r$. The direct path from start A to end B is just the diameter 2r$2r$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line

### Related Formula In rectilinear kinematics: - Velocity: v = fracdxdt$v = \frac{dx}{dt}$ - Acceleration: a = fracdvdt = fracd^2xdt^2$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$ ### Core Logic Given position-time function: $ ### Core Logic Given position-time function: $x(t) = c_0 (t^2 - 2) + c (t - 2)^2$x(t) = c_0 (t^2 - 2) + c (t - 2)^2$ ### Step 1: Differentiate once to get velocity ($ ### Step 1: Differentiate once to get velocity ($v) $) $v = fracdxdt = fracddtleft[c_0(t^2 - 2)right] + fracddtleft[c(t-2)^2right]$v = \frac{dx}{dt} = \frac{d}{dt}\left[c_0(t^2 - 2)\right] + \frac{d}{dt}\left[c(t-2)^2\right]$ $ $v = c_0 (2t) + c cdot 2(t-2) = 2 c_0 t + 2 c(t - 2)$v = c_0 (2t) + c \cdot 2(t-2) = 2 c_0 t + 2 c(t - 2)$ ### Step 2: Differentiate again to get acceleration ($ ### Step 2: Differentiate again to get acceleration ($a) $) $a = fracdvdt = fracddtleft[2 c_0 t + 2 c(t - 2)right]$a = \frac{dv}{dt} = \frac{d}{dt}\left[2 c_0 t + 2 c(t - 2)\right]$ $ $a = 2 c_0 + 2 c = 2(c + c_0)$a = 2 c_0 + 2 c = 2(c + c_0)$ This shows acceleration is constant and equals $ This shows acceleration is constant and equals $2(c + c_0), matching Statement (4). ### Pattern Recognition Whenever a position function is a pure quadratic polynomial in $, matching Statement (4). ### Pattern Recognition Whenever a position function is a pure quadratic polynomial in $t, the acceleration is constant and equal to $, the acceleration is constant and equal to $2 \times the coefficient of the $ the coefficient of the $t^2 term. Rewriting $ term. Rewriting $x(t): $: $x(t) = (c_0 + c)t^2 - 4ct + (4c - 2c_0)$x(t) = (c_0 + c)t^2 - 4ct + (4c - 2c_0)$ The coefficient of $ The coefficient of $t^2 is $ is $(c_0 + c). Thus, acceleration is $. Thus, acceleration is $2(c_0 + c)$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line

### Related Formula langle v rangle = fracDelta xDelta t = fracx_f - x_it_f - t_i$\langle v \rangle = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$ v_textinst = fracdxdt = textslope of xtext-t text graph$v_{\text{inst}} = \frac{dx}{dt} = \text{slope of } x\text{-}t \text{ graph}$ ### Core Logic Let's test each statement using coordinates from the given graph: - For (A): At t=0, x=0$t=0, x=0$; at t=3, x=5$t=3, x=5$. langle v rangle = frac5-03 = frac53text m/s neq 10text m/s$\langle v \rangle = \frac{5-0}{3} = \frac{5}{3}\text{ m/s} \neq 10\text{ m/s}$ (Incorrect). - For (B): At t=3, x=5$t=3, x=5$; at t=5, x=5$t=5, x=5$. langle v rangle = frac5-52 = 0text m/s$\langle v \rangle = \frac{5-5}{2} = 0\text{ m/s}$ (Correct). ### Step 1: Evaluate Remaining Statements - For (C): Segment from 0 to 3s passes through points (0, -5)$(0, -5)$ or starts linearly. The slope from t=0$t=0$ to t=3$t=3$ can be calculated from the linear line segment: textslope = frac5 - (-10)3 = 5text m/s$\text{slope} = \frac{5 - (-10)}{3} = 5\text{ m/s}$. Thus, instantaneous velocity at t=2text s$t=2\text{ s}$ is 5text m/s$5\text{ m/s}$ (Correct). - For (D): Slope during 5 to 7s vs instantaneous slope at t=6.5text s$t=6.5\text{ s}$ are completely different because the path changes slope. - For (E): At t=0, x=-5$t=0, x=-5$ and at t=9, x=-5$t=9, x=-5$. Since net displacement is zero, the average velocity from t=0$t=0$ to t=9text s$t=9\text{ s}$ is zero (Correct). Thus, (B), (C), and (E) are the correct statements. ### Pattern Recognition Average velocity requires only initial and final positions (x_f, x_i$x_f, x_i$). Instantaneous velocity reads directly off the segment's geometric slope. If initial and final coordinates match, average velocity is unconditionally zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line