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  4. Moving along the x-axis there are two points with x = 10 + 6t, x = 3 + t2. The speed with which they are reaching from each other at the time of encounter is (x is in cm and t is in seconds)

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Q. Moving along the $x$-axis there are two points with $x = 10 + 6t$, $x = 3 + t^2$. The speed with which they are reaching from each other at the time of encounter is ($x$ is in $cm$ and $t$ is in seconds)

Application of Derivatives

Solution:

They will encounter if
$10 + 6t = 3 + t^2$
$\Rightarrow t^{2}- 6t - 7 = 0$
$\Rightarrow t = 7$
At $t = 7\, s$, moving in a first point
$v_{1} = \frac{d}{dt}\left(10+6t\right) = 6\, cm/s$
At $t = 7s$, moving in a second point
$v_{2} = \frac{d}{dt}\left(3+t^{2}\right)$
$= 2t = 2 \times 7 = 14 \,cm/s$
$\therefore $ Resultant velocity $= v_{2} - v_{1} $
$= 14 - 6 = 8 \,cm/s$