Discussion
If the position of the particle as a function of time t is \(\vec{r} = 8t\hat{i} + 3t^{2}\hat{j} + 3\hat{k}\) m, then the acceleration of the particle is (in ms-2 )
Answer: A
The acceleration is 6 m s-2.
Differentiate the position twice:
\[ \begin{aligned} \vec{v} &= \frac{d\vec{r}}{dt} = 8\hat{i} + 6t\hat{j} \\ \vec{a} &= \frac{d\vec{v}}{dt} = 6\hat{j} \end{aligned} \]
The magnitude is 6. 3 is the coefficient of \(t^{2}\) without differentiating, and 8 is the constant velocity component.
The \(x\)-component \(8t\) gives a constant velocity and the \(z\)-component 3 is fixed, so only the \(y\)-motion is accelerated.
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