Moving along the x-axis are two points with x | vedclass

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(C) The points encounter each other when their positions are equal:$10 + 6t = 3 + t^2$Rearranging the terms gives the quadratic equation:$t^2 - 6t - 7

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$8$

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(C) The points encounter each other when their positions are equal:$10 + 6t = 3 + t^2$Rearranging the terms gives the quadratic equation:$t^2 - 6t - 7 = 0$$(t - 7)(t + 1) = 0$Since time $t$ cannot be negative,we have $t = 7 \, sec$.Now,we find the velocities of the two points by differentiating their position equations with respect to time $t$:$v_1 = \frac{dx_1}{dt} = \frac{d}{dt}(10 + 6t) = 6 \, cm/sec$$v_2 = \frac{dx_2}{dt} = \frac{d}{dt}(3 + t^2) = 2t$At $t = 7 \, sec$,the velocity of the second point is:$v_2 = 2 	imes 7 = 14 \, cm/sec$The relative speed at which they are receding from each other is the difference in their velocities:$|v_2 - v_1| = |14 - 6| = 8 \, cm/sec$.

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

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