At the moment $t = 0,$ a particle leaves the origin and moves in the positive direction of the x-axis. Its velocity varies with time as $v = v_{0} (1 - \frac{t}{5})$ where $v_{0} = 10 c m / \sec$ . The particle will be the distance of 10 cm from the origin at three different instants $T_{1}, T_{2}$ and $T_{3} [T_{3} > T_{2} > T_{1}]$ . If the value of $[T_{3} - T_{2}]$ is of the form $\frac{\sqrt{X} - \sqrt{Y}}{2}$ where X and Y are some numbers and $Y < X$ . Then the value of $[X - Y] / 10$ will be?

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answer is 8.

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Detailed Solution

$\frac{d x}{d t} = v_{0} - \frac{v_{0} t}{5} \Rightarrow x = v_{0} t - \frac{v_{0} t^{2}}{10}$
Where x can be either $+ 10 o r - 10$ .
$10 t - t^{2} = 10 \Rightarrow t = \frac{10 \pm \sqrt{60}}{2} 10 t - t^{2} = - 10 t = \frac{10 \pm \sqrt{140}}{2}$

For 2nd event, $t = \frac{10 + \sqrt{60}}{2}$
And for 3rd event, $t = \frac{10 + \sqrt{140}}{2}$