The position vector of a moving body at any instant of time is given as r = (5t² i - 5t j)m$\vec{\mathrm{r}} = (5\mathrm{t}^2\hat{\mathrm{i}} - 5\mathrm{t}\hat{\mathrm{j}})\mathrm{m}$. The magnitude and direction of velocity at t = 2s$t = 2s$ is,

Related Formula

Core Logic

To find the relative crossings, set the position functions equal: xP(t) = xQ(t)$x_P(t) = x_Q(t)$.

This is a cubic polynomial equation in terms of time t$t$. A cubic function can yield a maximum of 3 distinct real roots.

Depending on the initial velocity profiles (v0P, v0Q$v_{0P}, v_{0Q}$), the curves can intersect at t=0$t=0$ and potentially create up to two additional crossing paths downstream as acceleration profiles cross over. Thus, the maximum possible number of total crossings is exactly 3.

Pattern Recognition

Crossings correspond mathematically to intersections of relative displacement polynomials. A linear acceleration profile produces a cubic position curve (t³$t^3$), which can cross a quadratic curve (t²$t^2$) at up to 3 distinct coordinate points.

Chapter Mix

Class 11 Physics: Kinematics

Related Formula

For realistic physical motion in one dimension:

• Time t$t$ can never be negative during a normal positive time sequence, and cannot flow backwards.
• Total distance covered can never decrease over time.
• A particle cannot have two different values of position or velocity at the exact same instant of time.

Core Logic

Let us analyze each curve:

• Curve (A) (Phase φ$\phi$ vs Time t$t$): Represents φ = kt + C$\phi = kt + C$, which is a valid linear relationship of phase over time (e.g., in Simple Harmonic Motion x = A (kt + C)$x = A\sin(kt + C)$). (Valid)
• Curve (B) (Velocity v$v$ vs Displacement x$x$): A closed loop, which represents symmetric harmonic-type oscillation. For example, v² + ω² x² = const$v^2 + \omega^2 x^2 = \text{const}$ (ellipse) is a perfectly physically valid 1D SHM velocity-displacement phase portrait. (Valid)
• Curve (C) (Velocity vs Time): The curve enters into the negative time quadrant. Time cannot go backwards or exist in negative values relative to starting sequence in standard physical scenarios. (Invalid)
• Curve (D) (Total Distance vs Time): Represents total distance increasing over time. Total distance is a non-decreasing function of time (d(d)/dt ≥ 0$d(d)/dt \ge 0$). Thus, this curve is physically valid. (Valid)

Step 1: Conclusion

Therefore, curves A, B, and D possibly represent physical one-dimensional motion. The correct option is (1).

Pattern Recognition

Quick check for graph validity:

• Time cannot run backwards (ruling out C).
• Total distance can never decrease (D is valid because it strictly goes upwards).
• v$v$ vs x$x$ can be circular/elliptical in SHM (B is valid).

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Related Formula

Maximum height of a projectile:

where θ$\theta$ is the angle of projection measured from the horizontal ground.

Core Logic

The angle of projection in this question is measured from the vertical axis as φ$\phi$. This relates to the horizontal projection angle θ$\theta$ as:

Substitute this into the maximum height formula:

Step 1: Analyzing Boundary Conditions

Let's check the boundary values of hm$h_m$ for φ in [0, 90°]$\phi \in [0, 90^{\circ}]$:

• At φ = 0°$\phi = 0^{\circ}$ (fired straight up vertically):

• At φ = 90°$\phi = 90^{\circ}$ (fired horizontally along the ground):

Let's evaluate the behavior of hm(φ)$h_m(\phi)$:

• At small angles φ ≈ 0$\phi \approx 0$, the slope (d hm)/(dφ) = -(u²)/(2g) (2φ) ≈ 0$\frac{d h_m}{d\phi} = -\frac{u^2}{2g} \sin(2\phi) \approx 0$. This means the curve starts with a horizontal slope (flat top).
• As φ$\phi$ increases towards 90°$90^{\circ}$, hm$h_m$ falls smoothly, reaching 0 at 90°$90^{\circ}$ with flat slope again.

This sinusoidal falloff perfectly matches Curve (3).

Pattern Recognition

Always read the coordinate origin carefully! Measuring angle from the vertical instead of the horizontal turns ²θ$\sin^2\theta$ into ²φ$\cos^2\phi$. Checking extremes: vertical shot (0°$0^{\circ}$) yields maximum height, while flat shot (90°$90^{\circ}$) yields zero height.

Chapter Mix

Class 11 Physics: Motion in a Plane: Projectile Motion

Related Formula

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$t=3, x=5$. v = (5-0)/(3) = (5)/(3) m/s ≠ 10 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$t=5, x=5$. v = (5-5)/(2) = 0 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: slope = (5 - (-10))/(3) = 5 m/s$\text{slope} = \frac{5 - (-10)}{3} = 5\text{ m/s}$. Thus, instantaneous velocity at t=2 s$t=2\text{ s}$ is 5 m/s$5\text{ m/s}$ (Correct).
• For (D): Slope during 5 to 7s vs instantaneous slope at t=6.5 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=9 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 (xf, xᵢ$x_f, x_i$). Instantaneous velocity reads directly off the segment's geometric slope. If initial and final coordinates match, average velocity is unconditionally zero.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Related Formula

x = u · t$x = u \cdot t$ [cite: 763]

H = (1)/(2)gt²$$H = \frac{1}{2}gt^2$$ [cite: 764]

D = √(x² + H²)$$D = \sqrt{x^2 + H^2}$$ [cite: 774]

Core Logic

First, convert the horizontal velocity component to metric SI units: [cite: 157, 763]

u = 360 × (5)/(18) = 100 m/s$$u = 360 \times \frac{5}{18} = 100\ \text{m/s}$$ [cite: 157, 763]

Calculate the horizontal range distance x$x$ covered over t = 20 s$t = 20\ \text{s}$: [cite: 158, 763]

x = 100 × 20 = 2000 m = 2 km$$x = 100 \times 20 = 2000\ \text{m} = 2\ \text{km}$$ [cite: 158, 763]

The vertical displacement height H$H$ is explicitly given as 2 km = 2000 m$2\ \text{km} = 2000\ \text{m}$[cite: 157, 769]. Let's confirm with free-fall height calculation matching the solution template: [cite: 764]

H = (1)/(2) × 10 × (20)² = 5 × 400 = 2000 m = 2 km$$H = \frac{1}{2} \times 10 \times (20)^2 = 5 \times 400 = 2000\ \text{m} = 2\ \text{km}$$ [cite: 158, 769]

Now find the net spatial vector displacement D$D$ from the release coordinates: [cite: 774]

D = √(x² + H²) = √(2² + 2²) = √(8) = 2√(2) km$$D = \sqrt{x^2 + H^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\ \text{km}$$ [cite: 774]

Pattern Recognition

Be careful with the wording: the question asks for the displacement from the release coordinate position [cite: 159], which is the hypotenuse vector √(x² + H²)$\sqrt{x^2 + H^2}$[cite: 774]. Do not mistake it for the horizontal range distance alone.

Chapter Mix

Class 11 Physics: Motion in a Plane