Question

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)

• A. $16cm/s$
• B. $20cm/s$
• C. $8cm/s$
• D. $12cm/s$

Answer 1

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
$v_1=\frac{d}{dt}\left(10+6t\right)=6cm/s$
At $t=7s$, moving in a second point
$v_2=\frac{d}{dt}\left(3+t^2\right)$
$=2t=2\times 7=14cm/s$
$\therefore$ Resultant velocity $=v_2-v_1$
$=14-6=8cm/s$