The position vector of a particle of mass 3 kg moving along a space curve is given by \(r = (4t^{3} - t^{2})i - (2t^{2} - t)j\) at any time t seconds. Find ...
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The position vector of a particle of mass 3 kg moving along a space curve is given by \(r = (4t^{3} - t^{2})i - (2t^{2} - t)j\) at any time t seconds. Find the force acting on it at t = 2 seconds.
Answer Details
The force is found from Newton's second law \(\mathbf{F} = m\mathbf{a}\), where \(\mathbf{a}\) is the second derivative of the position vector.
Given \(\mathbf{r} = (4t^3 - t^2)\mathbf{i} - (2t^2 - t)\mathbf{j}\), differentiate once for velocity:
\[\mathbf{v} = \frac{d\mathbf{r}}{dt} = (12t^2 - 2t)\mathbf{i} - (4t - 1)\mathbf{j}.\]
Differentiate again for acceleration:
\[\mathbf{a} = \frac{d\mathbf{v}}{dt} = (24t - 2)\mathbf{i} - 4\mathbf{j}.\]
At \(t = 2\): \(\mathbf{a} = (24(2) - 2)\mathbf{i} - 4\mathbf{j} = 46\mathbf{i} - 4\mathbf{j}\).
With mass \(m = 3\ \text{kg}\):
\[\mathbf{F} = 3(46\mathbf{i} - 4\mathbf{j}) = 138\mathbf{i} - 12\mathbf{j}\ \text{N}.\]
Its magnitude is \(|\mathbf{F}| = \sqrt{138^2 + 12^2} = \sqrt{19188} \approx 138.5\ \text{N}\).
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