A particle starts from rest and moves in a straight line such that its velocity, v, at time t seconds is given by \(v = (3t^{2} - 2t) ms^{-1}\). Calculate t...
Question 1 Report
A particle starts from rest and moves in a straight line such that its velocity, v, at time t seconds is given by \(v = (3t^{2} - 2t) ms^{-1}\). Calculate the distance covered in the first 2 seconds.
Answer Details
To find the distance covered in the first 2 seconds, we need to integrate the velocity function from 0 to 2 seconds:
\begin{align*}
\text{Distance} &= \int_{0}^{2} v\,dt \\
&= \int_{0}^{2} (3t^2 - 2t)\,dt \\
&= \left[\frac{3}{3}t^3 - \frac{2}{2}t^2\right]_{0}^{2} \\
&= \left(3\cdot 2^2 - 2\cdot 2^2\right) - \left(3\cdot 0^2 - 2\cdot 0^2\right) \\
&= 4 \text{ m}.
\end{align*}
Therefore, the distance covered in the first 2 seconds is 4 meters. Answer choice (b) is correct.
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