Question

$x$ and $y$-coordinates of a particle in motion, as functions of time $t$ , are given by
$x=7t^2-4t+6,y=3t^3-3t^2-12t-5$ ( $x\text{and}y$ are in $m$ and $t$ is in $s$.)
The $x\text{and}y$-components of the average velocity, in the interval from $t=0\text{s to}t=5s$ are

• A. $v_x=32.2m{s}^{-1},v_y=47m{s}^{-1}$
• B. $v_x=-32.2m{s}^{-1},v_y=-47m{s}^{-1}$
• C. $v_x=31m{s}^{-1},v_y=48m{s}^{-1}$
• D. $v_x=-31{ms}^{-1},v_y=-48{ms}^{-1}$

Answer 1

The correct option is C $v_x=31m{s}^{-1},v_y=48m{s}^{-1}$

$x=7t^2-4t$

$y=3t^3-3t^2-12t-5$

$att=0$

$x=0,y=-5$

$att=5s$

$x=155,y=235$

$\therefore V{}_x=\frac{155}{5}=31m/s$

$V{}_y=\frac{235-\left(-5\right)}{5}=48m/s$

Hence,

option $\left(C\right)$ is correct answer.