We are given:
Step 1: Use Newton’s Second Law to find acceleration:
\[ \vec{a} = \frac{\vec{F}}{m} = 2t \hat{i} + 3t^2 \hat{j} \]
Step 2: Integrate to find velocity:
\[ \vec{v} = \int \vec{a} \, dt = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt = t^2 \hat{i} + t^3 \hat{j} \]
Step 3: Power is given by dot product \( \vec{F} \cdot \vec{v} \):
\[ P = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j}) = 2t^3 + 3t^5 \] \[ \Rightarrow P = 2t^3 + 3t^5 \, \text{W} \]
Given: \[ \vec{F} = (2t \hat{i} + 3t \hat{j}) \, \text{N} \] The mass of the object is \( m = 1000 \, \text{gm} = 1 \, \text{kg} \). Using Newton's second law: \[ \vec{F} = m \vec{a} \quad \Rightarrow \quad \vec{a} = 2t \hat{i} + 3t^2 \hat{j} \] The velocity is the integral of acceleration: \[ \frac{d\vec{v}}{dt} = 2t \hat{i} + 3t^2 \hat{j} \] Integrating with respect to time: \[ \vec{v} = t^2 \hat{i} + t^3 \hat{j} \] Now, the power \( P \) is given by the dot product of force and velocity: \[ P = \vec{F} \cdot \vec{v} \] Substitute the expressions for \( \vec{F} \) and \( \vec{v} \): \[ P = (2t \hat{i} + 3t \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j}) \] Simplifying: \[ P = (2t^3 + 3t^5) \, \text{W} \] \[ \boxed{P = 2t^3 + 3t^5 \, \text{W}} \]
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).

If a random variable \( x \) has the probability distribution 
then \( P(3<x \leq 6) \) is equal to