Distance with Linearly Increasing Acceleration – Rankers Physics
Topic: Kinematics
Subtopic: Equations of Motion

Distance with Linearly Increasing Acceleration

The acceleration of a particle is increasing linearly with time (t) as \(bt\). The particle starts from origin with an initial velocity \(v_0\). The distance travelled by the particle in time (t) will be:

(1995)

\(v_0 t + \frac{bt^2}{3}\)
\(v_0 t + \frac{bt^2}{2}\)
\(v_0 t + \frac{bt^3}{6}\)
\(v_0 t + \frac{bt^3}{3}\)

Solution:

Given \(a = \frac{dv}{dt} = bt\). Integrating, \(v = \int bt , dt = \frac{1}{2}bt^2 + C_1\). Since \(v=v_0\) at \(t=0\), \(C_1 = v_0\). So \(v = v_0 + \frac{1}{2}bt^2\). Given \(v = \frac{dx}{dt}\). Integrating again, \(x = \int (v_0 + \frac{1}{2}bt^2\) , \(dt = v_0 t + \frac{1}{2}b \frac{t^3}{3} + C_2\). Since (x=0) at (t=0), (C_2 = 0). Thus, \(x = v_0 t + \frac{bt^3}{6}\).

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