Question

A particle is moving along the x-axis with its coordinate with the time $t$ given by $x\left(t\right)=-3t^2+8t+10$meter. Another particle is moving along the y-axis with its coordinate as a function of time given by $y\left(t\right)=5-8t^3$meter. . At $t=1$ second , the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{v}$ . Then $v\left(m{s}^{-1}\right)$ is

Answer 1

Explanation for the correct option:

Step 1: According to the question:

Let particle A moves along X-axis such that

$x\left(t\right)=-3t^2+8t+10$

Let particle B moves along Y-axis such that

$y\left(t\right)=5-8t^3$

Step 2. Calculate the velocity of A and B:

The velocity of particles A

$\begin{array}{rcl}\overrightarrow{v_A}& =& v_x=\frac{dx}{dt}==(-6t+8)\\ or{\left(\overrightarrow{v_A}\right)}_{t=1}& =& (-6+8)\\ & =& 2m{s}^{-1}\\ & & \end{array}$

$\Rightarrow \overrightarrow{v_A}=+2\hat{i}m{s}^{-1}$

The velocity of particles B

$\begin{array}{rcl}\overrightarrow{v_B}& =& v_y=\frac{dy}{dt}=(-24t^2)\\ or{\left(\overrightarrow{v_B}\right)}_{t=1}& =& (-24)m{s}^{-1}\end{array}$

$\Rightarrow \overrightarrow{v_B}=-24\hat{j}m{s}^{-1}$

Speed of the particle B with respect to A,

$\Rightarrow \overrightarrow{v_{BA}}=(\overrightarrow{v_B}-\overrightarrow{v_A})=(-24\hat{j}-2\hat{i})=(-2\hat{i}-24\hat{j})$

Thus, the magnitude of the velocity $|\overrightarrow{v_{BA}}|=\sqrt{{\left(2\right)}^2+{\left(24\right)}^2}$

Step 3. Find the value of $v$ :

But, given

$$\begin{gathered} |\overrightarrow{v_{BA}}|=\sqrt{v} \\ \Rightarrow \sqrt{v}=\sqrt{580} \\ \Rightarrow v=580m{s}^{-1} \end{gathered}$$

Hence the velocity $v=580m{s}^{-1}$.