NEET Physics Question - Kinematics - Calculus Based Questions

Displacement (x) of a particle is related to time (t) as x = at + bt² – ct³, where a, b and c are constants of the motion. The velocity of the particle when its acceleration is zero is given by

\[a+\frac{b^{2}}{c}\]

\[a+\frac{b^{2}}{2c}\]

\[a+\frac{b^{2}}{3c}\]

\[a+\frac{b^{2}}{4c}\]

Solution:

Given:

\[
x = at + bt^2 - ct^3
\]

1. Velocity:

\[
v = \frac{dx}{dt} = a + 2bt - 3ct^2
\]

2.Acceleration:

\[
a = \frac{dv}{dt} = 2b - 6ct
\]

3. Set acceleration to zero:

\[
2b - 6ct = 0 ; t = \frac{b}{3c}
\]

4. Velocity at \(t = \frac{b}{3c}\)

\[
v = a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2
\]

\[
= a + \frac{2b^2}{3c} - \frac{b^2}{3c} = a + \frac{b^2}{3c}
\]

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Solution:

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