Velocity Time Graphs (Linear Motion) - Worked Example

1. A train, which starts its motion from rest, moves along a straight linear road. It  takes a velocity of $u$ by moving with an acceleration of $f_1$ and then a velocity of $3u$ by moving with an acceleration of $f_2$. After moving with this velocity for a period of $\cfrac{u}{f_1}$, it takes a deceleration of $f_2$ until it comes back to the rest. Show that the total displacement of the train is $\cfrac{u^2[17f_1+7f_2]}{2f_1 f_2}$.
Answer




On this graph, OA line segment represent the $f_1$ acceleration and AB represent the $f_2$ acceleration. At BC train moves with a constant velocity and then at CD it decelerate from $f_2$, as described in the question.
From the given data,
$\tan\alpha = f_1$
$\tan\beta = f_2$
OG $= u \cot\alpha = \cfrac{u}{f_1}$
GF $= (3u-u)\cot\beta = \cfrac{2u}{f_2}$
ED $= 3u \cot\beta = \cfrac{3u}{f_2}$

The total displacement = OABCD area
$\therefore$ Total displacement = $\cfrac{1}{2}$OG$\cdot$GA + $\cfrac{1}{2}$(AG+BF)$\cdot$GF + BF$\cdot$EF + $\cfrac{1}{2}$CE$\cdot$ED
                                    $= \cfrac{1}{2}\cdot u \cdot \cfrac{u}{f_1}+\cfrac{3u+u}{2}\cdot \cfrac{2u}{f_2}+3u\cdot \cfrac{u}{f_1}+\cfrac{1}{2}\cdot 3u \cdot \cfrac{3u}{f_2}$
                                    $= \cfrac{u^2}{2f_1}+\cfrac{8u^2}{2f_2}+\cfrac{3u^2}{f_1}+\cfrac{9u^2}{2f_2}$
                                    $=\cfrac{7u^2}{2f_1}+\cfrac{17u^2}{2f_2}$
                                    $=\cfrac{u^2[17f_1+7f_2]}{2f_1 f_2}$


2. To be continued...