a body with mass 1 kg.

Step 1: Find Acceleration

The acceleration is found by dividing the force by mass: $$\vec{a} = \frac{\vec{F}}{m} = (t \hat{i} + 2 t^2 \hat{j})$$, because mass $ m = 1 $ kg.

Step 2: Write the Relationship Between Acceleration and Velocity

Acceleration is the rate of change of velocity: $$\vec{a} = \frac{d\vec{v}}{dt}$$.

Step 3: Find the Velocity by Integration

To get velocity, integrate acceleration over time: $$ \vec{v} = \int \vec{a} \, dt = \int (t \hat{i} + 2 t^2 \hat{j}) dt $$

This gives: $$ \vec{v} = \frac{t^2}{2}\hat{i} + \frac{2 t^3}{3}\hat{j} $$

Step 4: Find the Power

Power is the dot product of force and velocity: $$ P = \vec{F} \cdot \vec{v} $$ So, $$ P = (t \hat{i} + 2 t^2 \hat{j}) \cdot \left(\frac{t^2}{2}\hat{i} + \frac{2 t^3}{3}\hat{j}\right) $$

Calculate the dot product:

$ t $ times $ \frac{t^2}{2} = \frac{t^3}{2} $ (from the x-components)
$ 2t^2 $ times $ \frac{2t^3}{3} = \frac{4t^5}{3} $ (from the y-components)

So, $$ P = \frac{t^3}{2} + \frac{4t^5}{3} $$

Step 5: Substitute $ t = 3 $ seconds

Put $ t = 3 $ into the formula: $$ P = \frac{3^3}{2} + \frac{4 \times 3^5}{3} $$ Calculate the values:

$ 3^3 = 27 $, so $ \frac{27}{2} = 13.5 $
$ 3^5 = 243 $, so $ 4 \times 243 = 972 $, and $ \frac{972}{3} = 324 $

Add them: $ 13.5 + 324 = 337.5 $

Final Answer:

Therefore, the power developed at $ t = 3 $ seconds is $ 337.5 \mathrm{~W} $.

Work Energy And Power

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