TS EAMCET · Physics · Motion In Two Dimensions
A particle moves in a circle with speed \(v\) varying with time as \(v(t)=2 t\). The total acceleration of the particle after it completes 2 rounds of cycle is
- A \(16 \pi\)
- B \(2 \sqrt{1+64 \pi^2}\)
- C \(2 \sqrt{1+49 \pi^2}\)
- D \(14 \pi\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{1+64 \pi^2}\)
Step-by-step Solution
Detailed explanation
Instantaneous velocity of particle in circular motion is \(v=r \omega=a t\), (let \(v=a t)\) Then, \(\quad \omega=\frac{d \theta}{d t}=\frac{a t}{r}\) So, \(\quad \int_0^{2 \pi n}=\int_0^t \frac{a t}{r} d t\), \(n=\) number of rounds \(=2\) (given)…
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