Discussion
If the velocity (in \(ms^{-1}\)) of a particle at any instant \(t\) is given by \(2.0\hat{i} + 3.0t\hat{j}\) then the magnitude of its acceleration (in \(ms^{-2}\)) is
Answer: B
The acceleration is 3 m s-2.
Acceleration is the time derivative of velocity:
\[ \begin{aligned} \vec{a} &= \frac{d}{dt}\left(2.0\hat{i} + 3.0t\hat{j}\right) \\ &= 3.0\hat{j} \end{aligned} \]
The constant \(x\)-component contributes nothing, so the magnitude is 3. Adding the coefficients 2 and 3 gives 5, and 2 is the constant velocity component, not an acceleration.
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