A particle is moving in a straight line. The variation of position ' x ' as a function of time ' t ' is given as x = (t³ - 6t² + 20t + 15) m. The velocity of the body when its acceleration becomes zero is: (a) 6 m/s (b) 10 m/s (c) 8 m/s (d) 4 m/s

Answer: (c). v = 3t² - 12t + 20, a = 6t - 12. a = 0 ⇒ t = 2 s. v = 12 - 24 + 20 = 8 m/s So the answer is (c) 8 m/s.

A particle is moving in a straight line. The variation of position '… | JEE Main 2024 (29 Jan, Shift 2)

Q8JEE Main 2024 · 29 JAN · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

A particle is moving in a straight line. The variation of position ' $x$ ' as a function of time ' $t$ ' is given as $x = \left(t^3 - 6t^2 + 20t + 15\right)\text{ m}$ . The velocity of the body when its acceleration becomes zero is:

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Answer: (c)

$v = 3t^2 - 12t + 20$ , $a = 6t - 12$ .
$a = 0 \Rightarrow t = 2$ s.
$v = 12 - 24 + 20 = 8$ m/s
So the answer is (c) 8 m/s.

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