The velocity (v) - Distance (x) graph is shown in figure. Which graph represents acceleration (a) versus distance (x) variation of this system?

Related Formula

Core Logic

Since drops fall at regular intervals, find the total time of fall for the first drop and divide it by the number of intervals to get the time gap between successive drops.

Step 1: Total Time of Fall

Time taken by the first drop to reach the ground (h = 5 ~m$h = 5 \mathrm{~m}$):

Step 2: Time Interval Between Drops

The first drop strikes the floor when the sixth drop begins to fall. There are exactly 5 intervals between the 1st and 6th drop. Time between each drop Δ t = 1 ~sec5 = 0.2 ~sec$\Delta t = \frac{1 \mathrm{~sec}}{5} = 0.2 \mathrm{~sec}$.

Step 3: Distance of the Fourth Drop

The 4th drop has been falling for 2 intervals (since drops 6, 5, 4 mean the 4th drop was released 2 × 0.2 ~s$2 \times 0.2 \mathrm{~s}$ after the 6th drop started... wait, no. The 4th drop was released before the 5th and 6th). Time of fall for the 4th$4^{\text{th}}$ drop is: It was released 2 intervals after the 1st drop, so it has been falling for 1.0 - 2(0.2) = 0.6 ~s$1.0 - 2(0.2) = 0.6 \mathrm{~s}$? Wait, the PDF solution says: "Time of fall for 4th$4^{\text{th}}$ drop is 1 - 0.6 = 0.4 ~sec$1 - 0.6 = 0.4 \mathrm{~sec}$" Let's trace: 6th drop (0s fall), 5th drop (0.2s fall), 4th drop (0.4s fall). Distance fallen by 4th$4^{\text{th}}$ drop:

Step 4: Height from Ground

Height from ground = Total height - Distance fallen

Pattern Recognition

In 'falling drops' problems, if N drops are in the air, there are (N-1) time intervals. The kth$k^{\text{th}}$ drop from the top has fallen for (k-1)$(k-1)$ intervals.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Related Formula

Core Logic

Given x(t) = 4t³ - 3t$x(t) = 4t^3 - 3t$. The particle returns to origin when x = 0$x = 0$:

t = 0$t = 0$ (start) and t = √((3)/(4)) = √(3)2 ≈ 0.866 s$t = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.866 \text{ s}$. Thus, statement A is correct.

Step 1: Velocity and Turning Point

Velocity is v = (dx)/(dt) = 12t² - 3$v = \frac{dx}{dt} = 12t^2 - 3$. At turning point, velocity becomes zero:

(We take t > 0$t > 0$ since motion starts at t=0$t=0$).

Step 2: Position at Turning Point

Substitute t = 1/2$t = 1/2$ into x(t)$x(t)$:

The particle is |-1| = 1$|-1| = 1$ unit away from origin. Statement B is correct, and D is incorrect.

Step 3: Acceleration check

Acceleration a = (dv)/(dt) = 24t$a = \frac{dv}{dt} = 24t$. Since t ≥ 0$t \geq 0$, a ≥ 0$a \geq 0$, so acceleration is always non-negative. Statement C is correct. Statement E is incorrect because the particle does turn back (at t = 0.5$t = 0.5$, velocity changes sign from negative to positive).

Step 4: Final Conclusion

Statements A, B, and C are correct.

Pattern Recognition

To analyze 1D motion, simply find roots of x(t)=0$x(t)=0$ (origin passes), v(t)=0$v(t)=0$ (turning points), and check the sign of a(t)$a(t)$ over the given domain t ≥ 0$t \geq 0$.

Chapter Mix

Class 11 Physics: Kinematics

Related Formula

• Distance: Total actual path length covered.
• Displacement: Shortest straight-line distance connecting the initial and final position.
• Circumference of a complete circle = 2π r$2\pi r$
• Semicircular arc length = π r$\pi r$

Core Logic

The trajectory is defined by the sequence of points A → B → A → B$A \to B \to A \to B$:

• Distance Travelled:
• Segment 1 (A → B$A \to B$): Semicircular path of length π r$\pi r$
• Segment 2 (B → A$B \to A$): Semicircular path of length π r$\pi r$
• Segment 3 (A → B$A \to B$): Semicircular path of length π r$\pi r$
• Total distance:

• 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:

Step 1: Write result

The actual distance travelled is 3π 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 × π r = 3π r$3 \times \pi r = 3\pi r$. The direct path from start A to end B is just the diameter 2r$2r$.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Related Formula

Core Logic

Let the West-to-East river flow direction be along the positive x$x$-axis. The boat's velocity relative to water is vbr = 27~km/h$v_{br} = 27\mathrm{~km/h}$ at an angle of 150°$150^{\circ}$ with respect to the river flow (which is 30°$30^{\circ}$ upstream from the perpendicular crossing line).

To find the width of the river, we only need the component of the boat's velocity perpendicular to the river bank (y$y$-axis):

Convert this velocity to SI units:

Given the crossing time is half a minute (t = 30~s$t = 30\mathrm{~s}$):

Step 1: Final Conclusion

The width of the river is 112.5~m$112.5\mathrm{~m}$.

Pattern Recognition

In river-boat crossing problems, river velocity (vᵣ$v_r$) only causes drift along the bank; it has zero impact on crossing time or the crossing width calculations when the angle is defined relative to the flow. Use the vertical velocity component exclusively.

Chapter Mix

Class 11 Physics: Kinematics