Related Formula
F = m a$$\vec{F} = m\vec{a}$$
a = d vdt v = ∫ a dt$$\vec{a} = \frac{d\vec{v}}{dt} \implies \vec{v} = \int \vec{a} dt$$
P = F · v$$P = \vec{F} \cdot \vec{v}$$
Core Logic
First find acceleration from force and mass. Then integrate acceleration to find the velocity vector at time t$t$ (starting from rest). Finally, compute the dot product of Force and Velocity to get instantaneous power.
Step 1: Calculate Acceleration
Given F = (6t i + 6t² j) N$\vec{F} = (6t\hat{i} + 6t^2\hat{j}) \text{ N}$ and m = 2 kg$m = 2 \text{ kg}$.
a = Fm = 6t i + 6t² j2$$\vec{a} = \frac{\vec{F}}{m} = \frac{6t\hat{i} + 6t^2\hat{j}}{2}$$
a = (3t i + 3t² j) m/s²$$\vec{a} = (3t\hat{i} + 3t^2\hat{j}) \text{ m/s}^2$$
Step 2: Calculate Velocity
Assuming the body begins to move from rest (at t=0, v=0$t=0, v=0$):
v = ∫₀^t a dt = ∫₀^t (3t i + 3t² j) dt$$\vec{v} = \int_0^t \vec{a} \, dt = \int_0^t (3t\hat{i} + 3t^2\hat{j}) \, dt$$
v = ( (3t²)/(2) ) i + ( (3t³)/(3) ) j$$\vec{v} = \left( \frac{3t^2}{2} \right)\hat{i} + \left( \frac{3t^3}{3} \right)\hat{j}$$
v = ( (3t²)/(2) ) i + t³ j$$\vec{v} = \left( \frac{3t^2}{2} \right)\hat{i} + t^3\hat{j}$$
Step 3: Calculate Power
P = F · v$$P = \vec{F} \cdot \vec{v}$$
P = (6t i + 6t² j) · ((3t²)/(2) i + t³ j)$$P = (6t\hat{i} + 6t^2\hat{j}) \cdot \left(\frac{3t^2}{2}\hat{i} + t^3\hat{j}\right)$$
P = (6t) × ((3t²)/(2)) + (6t²) × (t³)$$P = (6t) \times \left(\frac{3t^2}{2}\right) + (6t^2) \times (t^3)$$
P = 9t³ + 6t⁵ W$$P = 9t^3 + 6t^5 \text{ W}$$
Pattern Recognition
When force varies as a polynomial in time tⁿ$t^n$, acceleration does too. Velocity jumps to tⁿ⁺¹$t^{n+1}$. Power (F · v$F \cdot v$) will result in terms behaving as t²ⁿ⁺¹$t^{2n+1}$. Here t → t³$t \to t^3$ and t² → t⁵$t^2 \to t^5$.
Chapter Mix
Class 11 Physics: Work, Energy and Power
Class 11 Physics: Motion in a Plane
