KCET · Maths · Application of Derivatives
A particle starts form rest and its angular displacement (in radians) is given by \(\theta=\frac{t^{2}}{20}+\frac{t}{5}\). If the angular velocity at the end of \(t=4\) is \(k\), then the value of \(5 k\) is
- A \(0.6\)
- B 5
- C \(5 k\)
- D 3
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
\(\theta=\frac{t^{2}}{20}+\frac{t}{5}\)
On differentiating \(\theta\) w.r.t. \(t\),
\(\begin{aligned} \frac{d \theta}{d t} &=\frac{t}{10}+\frac{1}{5} \\\left(\frac{d \theta}{d t}\right)_{t=4} &=\frac{4}{10}+\frac{1}{5} \end{aligned}\)
\(\begin{aligned} \Rightarrow & k &=\frac{3}{5} \\ \therefore & & 5 k &=3 \end{aligned}\)
On differentiating \(\theta\) w.r.t. \(t\),
\(\begin{aligned} \frac{d \theta}{d t} &=\frac{t}{10}+\frac{1}{5} \\\left(\frac{d \theta}{d t}\right)_{t=4} &=\frac{4}{10}+\frac{1}{5} \end{aligned}\)
\(\begin{aligned} \Rightarrow & k &=\frac{3}{5} \\ \therefore & & 5 k &=3 \end{aligned}\)
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