The displacement x$x$ versus time graph is shown below.

(A) The average velocity during 0 to 3 s is 10 m/s (B) The average velocity during 3 to 5 s is 0 m/s (C) The instantaneous velocity at t=2$t=2$ s is 5 m/s (D) The average velocity during 5 to 7 s and instantaneous velocity at t=6.5$t=6.5$ s are equal (E) The average velocity from t=0$t=0$ to t=9$t=9$ s is zero Choose the correct answer from the options given below:

Related Formula

For a body starting from rest (u = 0$u = 0$) with constant acceleration a$a$, displacement S$S$ in time t$t$ is:

Core Logic

Let the constant acceleration be a$a$.

• Displacement S₁$S_1$ covered in the first (p - 1)$(p - 1)$ seconds:

• Displacement S₂$S_2$ covered in the first p$p$ seconds:

Step 1: Express Net Displacement

Let the time taken to achieve a displacement of S₁ + S₂$S_1 + S_2$ be t$t$.

Substituting the expressions of S₁$S_1$ and S₂$S_2$:

Step 2: Solve for Time

Dividing the entire equation by (1)/(2)a$\frac{1}{2}a$:

Pattern Recognition

When dealing with equations of motion from rest, notice that displacement scales quadratically with time (S ∝ t²$S \propto t^2$). This means if displacement sums up (Sₜ = S₁ + S₂$S_t = S_1 + S_2$), the corresponding times will add in quadrature: t = √(t₁² + t₂²)$t = \sqrt{t_1^2 + t_2^2}$.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Related Formula

The centripetal force (F$F$) acting on a particle of mass m$m$ moving with velocity v$v$ in a path of radius r$r$ is given by:

Core Logic

Given that the two particles have the same mass (m₁ = m₂$m_1 = m_2$) and the ratio of their radii of curvature is:

To maintain a constant (equal) centripetal force (F₁ = F₂$F_1 = F_2$):

Step 1: Relate Velocity to Radius

Since m₁ = m₂$m_1 = m_2$, the equation simplifies to:

Step 2: Calculate the Ratio

Substituting the given ratio of radii:

Therefore, the ratio of their velocities is √(3):2$\sqrt{3}:2$.

Pattern Recognition

For problems involving steady values under constraint variations, establish the proportionality relation first. Here, F ∝ (v²)/(r) v ∝ √(r)$F \propto \frac{v^2}{r} \implies v \propto \sqrt{r}$ when F$F$ and m$m$ are held constant.

Chapter Mix

Class 11 Physics: Motion in a Plane

Core Logic

For two projectiles launched from the same height to have the same time of flight (T$T$), their vertical components of velocity must be equal. Since the angles given are with the vertical, the vertical components are vA (45°)$v_A \cos(45^{\circ})$ and vB (60°)$v_B \cos(60^{\circ})$. If TA = TB$T_A = T_B$, then vAy = vBy$v_{Ay} = v_{By}$.

Step 1: Check Inconsistency

For the ranges to be the same while having the same time of flight, their horizontal components of velocity must also be equal: vAx = vBx$v_{Ax} = v_{Bx}$.

This yields a contradiction. It is impossible for both the ranges and the times of flight to be simultaneously equal for different angles of projection from a tower.

Step 2: Conclusion

The question contains inconsistent data and is technically a Bonus question. However, if one arbitrarily equates only the time of flight (or if NTA intended a different scenario), the ratio (vA)/(vB) = 1√(2)$\frac{v_A}{v_B} = \frac{1}{\sqrt{2}}$ matches option (4), which was the officially provided key before corrections.

Pattern Recognition

Be wary of over-constrained physics problems. If a question specifies both range AND time of flight are identical for two different angles, check if the math yields a contradiction. NTA often accepts the result of one partial constraint By NTA 4 BY Rankbit (Bonus).

Chapter Mix

Class 11 Physics: Motion in a Plane

Related Formula

Core Logic

We need to find a new vector N$\vec{N}$ that has the magnitude of A$\vec{A}$ and the direction of B$\vec{B}$. Magnitude of A$\vec{A}$: | A| = √(3² + 4²) = √(25) = 5$|\vec{A}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$. Unit vector in the direction of B$\vec{B}$: B = B| B| = 4 i + 3 j√(4² + 3²) = 4 i + 3 j5$\hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{4\hat{i} + 3\hat{j}}{\sqrt{4^2 + 3^2}} = \frac{4\hat{i} + 3\hat{j}}{5}$.

Step 1: Construct the Vector

Step 2: Match Components

The x$x$ and y$y$ components are given as x$x$ and 3$3$. From N = 4 i + 3 j$\vec{N} = 4\hat{i} + 3\hat{j}$, we see the x$x$-component is 4$4$. Therefore, x = 4$x = 4$.

Pattern Recognition

Constructing a vector matching magnitude and direction is a simple scalar multiplication of the desired magnitude by the target direction's unit vector.

Chapter Mix

Class 11 Physics: Motion in a Plane

Related Formula

v = u + at$v = u + at$

Core Logic

Let the velocity at time t$t$ be u$u$. The time interval Δ t = (t+1) - t = 1 ~s$\Delta t = (t+1) - t = 1 \mathrm{~s}$. The increase in velocity over 1 second is exactly the acceleration a$a$. The displacement in that 1-second interval acts as the (t+1)th$(t+1)^{\text{th}}$ second displacement equation.

Step 1: Determine Acceleration

Increase in velocity Δ v = 50 ~m/s$\Delta v = 50 \mathrm{~m/s}$ in 1 ~s$1 \mathrm{~s}$. v = u + at$v = u + at$

Step 2: Determine Velocity 'u' at time t

Displacement in the 1-second interval from t$t$ to t+1$t+1$ is 125 ~m$125 \mathrm{~m}$. Using s = ut' + (1)/(2)at'²$s = ut' + \frac{1}{2}at'^2$ where t' = 1 ~s$t' = 1 \mathrm{~s}$:

Step 3: Distance in the next second

We need the distance travelled in the (t+2)th$(t+2)^{\text{th}}$ second, which corresponds to the 1-second interval starting with an initial velocity equal to the velocity at t+1$t+1$. Alternatively, we can use the nth$n^{\text{th}}$ second formula directly by re-indexing. The velocity at start of this interval is unew = u + a = 100 + 50 = 150 ~m/s$u_{new} = u + a = 100 + 50 = 150 \mathrm{~m/s}$. Distance S₁ₛₜ$S_{1\text{st}}$ using new parameters:

Or using PDF logic from base u$u$ with n=2$n=2$ (since n=1$n=1$ was t+1th$t+1^{\text{th}}$):

Pattern Recognition

For 1-second interval mechanics, Δ v$\Delta v$ directly yields a$a$. The displacement equation simplifies gracefully to s = uₛₜₐᵣₜ + a/2$s = u_{start} + a/2$.

Chapter Mix

Class 11 Physics: Motion in a Straight Line