MHT CET · Maths · Differentiation
The velocity of a particle at time \(t\) is given by the relation \(v=6 t-\frac{t^{2}}{6}\). The distance traveled in \(3 \mathrm{~s}\) is, if \(s=0\) at \(t=0\)
- A \(\frac{39}{2}\)
- B \(\frac{57}{2}\)
- C \(\frac{51}{2}\)
- D \(\frac{33}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{51}{2}\)
Step-by-step Solution
Detailed explanation
Given, \(v=\frac{d s}{d t}=6 t-\frac{t^{2}}{6}\)
On integrating both sides, we get
\(
s=3 t^{2}-\frac{t^{3}}{18}+\text { constant }
\)
Now, put \(s=0\) at \(t=0\), we get constant \(=0\)
\(
\therefore s=3 t^{2}-\frac{t^{3}}{18}
\)
Now, distance traveled in \(3 s=3(3)^{2}-\frac{(3)^{3}}{18}\)
\(
=27-\frac{27}{18}=\frac{51}{2}
\)
On integrating both sides, we get
\(
s=3 t^{2}-\frac{t^{3}}{18}+\text { constant }
\)
Now, put \(s=0\) at \(t=0\), we get constant \(=0\)
\(
\therefore s=3 t^{2}-\frac{t^{3}}{18}
\)
Now, distance traveled in \(3 s=3(3)^{2}-\frac{(3)^{3}}{18}\)
\(
=27-\frac{27}{18}=\frac{51}{2}
\)
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