The velocity of an object moving in a straight line path is given as a function of time by \(v=6 t-3 t^2\),…
The velocity of an object moving in a straight line path is given as a function of time by \(v=6 t-3 t^2\), where \(v\) is in \(\mathrm{ms}^{-1}, t\) is in \(\mathrm{s}\). The average velocity of the object between, \(t=0\) and \(t=2 \mathrm{~s}\) is
0
\(3 \mathrm{~ms}^{-1}\)
\(2 \mathrm{~ms}^{-1}\)
\(4 \mathrm{~ms}^{-1}\)
Solution
Given, velocity, \(v=6 t-3 t^2\)
As we know that,
\(v=\frac{d x}{d t}\)
Here, \(x\) is the displacement of the particle.
Now, \(d x=v d t\)
Integrate on the both sides, limit \(t=0\) to \(t=2\), we get
\(\begin{aligned} \therefore \quad x & =\int_0^2 v d t=\int_0^2\left(6 t-3 t^2\right) d t \\ & =\left[\frac{6 t^2}{2}\right]_0^2-\left[\frac{3 t^3}{3}\right]_0^2=\left[3 t^2\right]_0^2-\left[t^3\right]_0^2 \\ & =\left[3(2)^2-3(0)^2\right]-\left[(2)^3-(0)^2\right] \\ & =[12-0]-[8-0]=12-8=4 \mathrm{~m}\end{aligned}\)
\(\begin{aligned}\text { Average velocity, }
v_{\text {avg }} & =\frac{\text { Total displacement }}{\text { Total time taken }} \\
& =\frac{4}{2}=2 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
Hence, the average velocity of the object between \(t=0\) to \(t=2 \mathrm{~s}\) is \(2 \mathrm{~m} / \mathrm{s}\).