Question

Moving along the X-Axis are two points with $x=10+6t$ and $x=3+t^2$ The Speed with which they are reaching from each other at the time of encounter is ( $x$ is in $cm$ and $t$ is in $sec$ )

• A. $16cm.se{c}^{-1}$
• B. $20cm.se{c}^{-1}$
• C. $8cm.se{c}^{-1}$
• D. $12cm.se{c}^{-1}$

Answer 1

The correct option is C

$8cm.se{c}^{-1}$

Explanation for the correct answer:

Given that the Position of the two moving Points are $x=10+6t$ and $x=3+t^2$ In order to happen encounter between them their position will be the same such that $10+6t=3+t^2$

$\Rightarrow t^2-6t-7=0$

$\Rightarrow \left(t-7\right)\left(t+1\right)=0$

Since, Time can not be negative So, at $t=7$ encounter will happen between two points.

Now, Speed is the derivative of distance so, speed for the first point will be

$v_{Point1}=\frac{d\left(10+6t\right)}{dt}$

$v_{Point1}=6cm.se{c}^{-1}$

Speed for the second Point Will be

$v_{Point2}=\frac{d\left(3+t^2\right)}{dt}$

$v_{Point2}=2t$ Putting $t=7$ we get,

$v_{Point2}=14cm.se{c}^{-1}$

So, The resultant velocity of two Points will be

$$\begin{gathered} =v_{Point2}-v_{Point1} \\ =14-6 \\ =8cm.se{c}^{-1} \end{gathered}$$

Hence, the correct answer is option (C)