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 in cm and t is in seconds)

Question

Dr. Abubacker · Accepted Answer

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

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 in cm and t is in seconds)

Explanation

Reviewed by Dr. Abubacker

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