Question

the speed of a train increases at a constant rate of A from zero to V,and then remains constant for an interval ,and finally decreases to zero at a constant rate B.if L be the total distance travelled,then find the total time taken.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{when}\mathrm{velocity}\mathrm{increases}\mathrm{from}0\mathrm{to}V \\ \mathrm{Initial}\mathrm{velocity},u=0 \\ \mathrm{Final}\mathrm{velocity},v=V \\ \mathrm{Acceleration},a=A \\ \mathrm{Using} \\ v=u+at \\ \mathit{\Rightarrow }t\mathit{=}\frac{V}{A} \\ \mathrm{Using}\mathrm{the}\mathrm{second}\mathrm{equation}\mathrm{of}\mathrm{motion} \\ S=ut+\frac{1}{2}a{t}^{\mathit{2}} \\ \mathit{\Rightarrow }S=\frac{1}{2}A\times \frac{V^2}{A^2}=\frac{V^{\mathit{2}}}{2A} \\ \left(\mathrm{ii}\right)\mathrm{When}\mathrm{velocity}\mathrm{decreases}\mathrm{from}V\mathrm{to}0 \\ \mathrm{Initial}\mathrm{velocity},u=V \\ \mathrm{Acceleration},a=-B \\ \mathrm{Using} \\ v=u+at \\ \Rightarrow 0=\mathrm{V}-Bt \\ \mathit{\Rightarrow }t=\frac{V}{B} \\ S=ut+\frac{1}{2}a{t}^2 \\ \mathit{\Rightarrow }S=V\left(\frac{V}{B}\right)+\frac{1}{2}\left(-\mathrm{B}\right)\times {\left(\frac{\mathrm{V}}{\mathrm{B}}\right)}^2 \\ \Rightarrow S=\frac{V^2}{B}-\frac{V^2}{2B}=\frac{V^{\mathit{2}}}{2B} \\ \mathrm{So},\mathrm{The}\mathrm{total}\mathrm{distance}\mathrm{travelled}=\frac{V^2}{2}\left(\frac{1}{A}+\frac{1}{B}\right) \\ \left(\mathrm{iii}\right)\mathrm{Thus},\mathrm{distance}\mathrm{traveled}\mathrm{during}\mathrm{constant}\mathrm{velocity}\mathrm{phase} \\ =L-\frac{V^{\mathit{2}}}{2}\left(\frac{1}{A}+\frac{1}{B}\right) \\ \mathrm{So},\mathrm{time}\mathrm{taken}\mathrm{will}\mathrm{be} \\ T=\frac{L-{\displaystyle \frac{V^2}{2}}\left({\displaystyle \frac{1}{A}+\frac{1}{B}}\right)}{V} \\ \Rightarrow T=\frac{L}{V}-\frac{V}{2}\left(\frac{1}{A}+\frac{1}{B}\right) \\ \mathrm{So},\mathrm{total}\mathrm{time}\mathrm{to}\mathrm{cover}\mathrm{the}\mathrm{distance}L\mathrm{will}\mathrm{be} \\ =\left(\frac{V}{A}\right)+\left(\frac{L}{V}-\frac{V}{2}\left(\frac{1}{A}+\frac{1}{B}\right)\right)+\left(\frac{V}{B}\right) \\ =\frac{L}{V}+\frac{V}{A}-\frac{V}{2A}+\frac{V}{B}-\frac{V}{2B} \\ =\frac{\mathrm{L}}{\mathrm{V}}+\frac{\mathrm{V}}{2}\left(\frac{1}{\mathrm{A}}+\frac{1}{B}\right) \end{gathered}$$