TS EAMCET · Physics · Mathematics in Physics
A particle moves in \(X Y\)-plane with \(x\) and \(y\) varying with time \(t\) as \(x(t)=5 t, y(t)=5 t\left(27-t^2\right)\). At what time in seconds, the direction of velocity and acceleration will be perpendicular to each other?
- A \(5 \sqrt{\frac{27}{2}}\)
- B 5
- C \(5 \sqrt{12}\)
- D 3
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
Velocity in \(x\)-direction is \(v_x=\frac{d}{d t}(x)=\frac{d}{d t}(5 t)=5 \mathrm{~ms}^{-1}\) Velocity in \(y\)-direction is \(v_y=\frac{d}{d t}(y)=\frac{d}{d t}\left(5 \times 27 t-5 t^3\right)=5 \times 27-5 \times 3 t^2\) As, \(v_x=\) constant and \(v_y=\) time dependent.…
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