A particle moves along the x-axis and has its displacement x varying with time t according to the equation x=c_0(t^2-2)+c(t-2)^2$x=c_{0}(t^{2}-2)+c(t-2)^{2}$ where c_0$c_{0}$ and c are constants of appropriate dimensions. Then, which of the following statements is correct?

### 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 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

### Related Formula The distance traveled by an object equals the total area enclosed under its velocity-time (v-t$v-t$) plot along the time axis, treating all regional boundaries as strictly positive metrics: textDistance = int |v| \, dt$$\text{Distance} = \int |v| \, dt$$ ### Core Logic From the geometric grid profile between t = 0$t = 0$ and t = 4text s$t = 4\text{ s}$: 1. **First Region (t=0$t=0$ to t=2text s$t=2\text{ s}$)**: Forms a \right-angled \triangle with base = 2text s$= 2\text{ s}$ and peak height = 10text ms^-1$= 10\text{ ms}^{-1}$. textArea_1 = frac12 times 2 times 10 = 10 text m$$\text{Area}_1 = \frac{1}{2} \times 2 \times 10 = 10 \text{ m}$$ 2. **Second Region (t=2$t=2$ to t=4text s$t=4\text{ s}$)**: Forms a standard rectangle with width = (4 - 2) = 2text s$= (4 - 2) = 2\text{ s}$ and height = 10text ms^-1$= 10\text{ ms}^{-1}$. textArea_2 = 2 times 10 = 20 text m$$\text{Area}_2 = 2 \times 10 = 20 \text{ m}$$ Summing the areas together to extract total displacement path: textTotal Distance = 10 + 20 = 30 text m$$\text{Total Distance} = 10 + 20 = 30 \text{ m}$$ ### Pattern Recognition Always differentiate between distance and displacement on graph tracks. Displacement treats components below the axis as negative fields, while distance calculates absolute geometric magnitudes without direction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line

### Related Formula Velocity via first derivative of position: v = fracdxdt$$v = \frac{dx}{dt}$$ Acceleration via derivative of velocity: a = fracdvdt = fracd^2xdt^2$$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$$ ### Core Logic Given x(t) = -3t^3 + 18t^2 + 16t$x(t) = -3t^3 + 18t^2 + 16t$. Differentiate once to find velocity v(t)$v(t)$: v = fracdxdt = -9t^2 + 36t + 16$$v = \frac{dx}{dt} = -9t^2 + 36t + 16$$ Differentiate again to find acceleration a(t)$a(t)$: a = fracdvdt = -18t + 36$$a = \frac{dv}{dt} = -18t + 36$$ ### Step 1: Find the Time When Acceleration is Zero Set the acceleration equation to zero: -18t + 36 = 0 implies 18t = 36 implies t = 2mathrm~s$$-18t + 36 = 0 \implies 18t = 36 \implies t = 2\mathrm{~s}$$ ### Step 2: Calculate Velocity at Target Time Substitute t = 2mathrm~s$t = 2\mathrm{~s}$ back into the velocity function: v(2) = -9(2)^2 + 36(2) + 16$$v(2) = -9(2)^2 + 36(2) + 16$$ v(2) = -36 + 72 + 16 = 36 + 16 = 52mathrm~ms^-1$$v(2) = -36 + 72 + 16 = 36 + 16 = 52\mathrm{~ms}^{-1}$$ *Note on original structural text step matching:* The original calculation step text features a minor print truncation (v = -9t^2 + 36 + 16$v = -9t^2 + 36 + 16$), but explicitly evaluates out to the correct total of 52mathrm~m/s$52\mathrm{~m/s}$. ### Pattern Recognition Inflection point tracking: Finding where acceleration equals zero is mathematically identical to finding the maximum velocity point on the curve. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Straight Line