JEE Mains · Physics · STD 11 - 2. motion in straight line
A particle is moving in one dimension (along \(\mathrm{x}\) axis) under the action of a variable force. It's initial position was \(16 \mathrm{~m}\) right of origin. The variation of its position ( \(\mathrm{x}\) ) with time ( \(\mathrm{t})\) is given as \(\mathrm{x}=-3 \mathrm{t}^3+18 \mathrm{t}^2+16 \mathrm{t}\), where \(\mathrm{x}\) is in \(\mathrm{m}\) and \(\mathrm{t}\) is in \(\mathrm{s}\). The velocity of the particle when its acceleration becomes zero is _______ \(\mathrm{m} / \mathrm{s}.\)
- A \(50\)
- B \(52\)
- C \(57\)
- D \(60\)
Answer & Solution
Correct Answer
(B) \(52\)
Step-by-step Solution
Detailed explanation
\(x=3 t^3+18 t^2+16 t\) \(v=-9 \mathrm{t}^2+36t+16\) \(a=-18 \mathrm{t}+36\) \(a=0 \text { at } \mathrm{t}=2 \mathrm{~s}\) \(v (2)=-9(2)^2+36 \times 2+16\) \(v=52 \mathrm{~m} / \mathrm{s}\)
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