STANDARD Physics Velocity and Acceleration MCQ Question
The position of a particle as a function of time t, is given by x(t) = at + bt² − ct³ where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be
a + b²/4c
a + b²/3c
a + b²/c
a + b²/2c
Correct Answer
Detailed Explanation
Here x(t) = at + bt² − ct³. ∴ v(t) = dx(t)/dt = a + 2bt − 3ct²; a(t) = dv(t)/dt = 2b − 6ct. ∴ a(t) = 0 ⇒ 2b − 6ct = 0 ⇒ t = 2b/6c. ∴ v = a + 2b(2b/6c) − 3c(2b/6c)² = a + b²/3c.
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