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$ v(t) = \begin{cases} \alpha t, & 0 \leq t \leq \frac{T}{3} \\ \frac{\alpha T}{3}, & \frac{T}{3} \leq t \leq \frac{2T}{3} \\ \alpha(T-t), & \frac{2T}{3} \leq t \leq T \end{cases} $

Where $\alpha \neq 0$ is a constant. To find the total displacement of the particle from time $ t = 0 $ to $ t = T $, we use the integral of velocity to find displacement, $ s = \int v(t) \, dt $.

Displacement Calculation:

For $ 0 \leq t \leq \frac{T}{3} $:

$ s_1 = \int_0^{\frac{T}{3}} \alpha t \, dt = \left[\frac{\alpha t^2}{2}\right]_0^{\frac{T}{3}} = \frac{\alpha}{2} \left(\frac{T^2}{9}\right) = \frac{\alpha T^2}{18} $

For $ \frac{T}{3} \leq t \leq \frac{2T}{3} $:

$ s_2 = \int_{\frac{T}{3}}^{\frac{2T}{3}} \frac{\alpha T}{3} \, dt = \left[\frac{\alpha T t}{3}\right]_{\frac{T}{3}}^{\frac{2T}{3}} = \frac{\alpha T}{9} $

For $ \frac{2T}{3} \leq t \leq T $:

$ s_3 = \int_{\frac{2T}{3}}^T \alpha(T-t) \, dt = \left[\alpha(Tt - \frac{t^2}{2})\right]_{\frac{2T}{3}}^T $

Simplifying further:

$ = \alpha \left[T \left(\frac{T}{3}\right) - \frac{1}{2}\left(T^2 - \frac{4T^2}{9}\right) \right] = \frac{\alpha T^2}{18} $

Total Displacement:

Adding all the displacements:

$ s_{\text{net}} = s_1 + s_2 + s_3 = \frac{\alpha T^2}{18} + \frac{\alpha T}{9} + \frac{\alpha T^2}{18} = \frac{2 \alpha T^2}{9} $

Motion In A Straight Line

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