Question

The position coordinates of a particle moving in $X-Y$ as a function of time $t$ are

$x=2t^{2}+6t+25$

$y=t^{2}+2t+1$

The speed of the object at $t=10 \, s$ is approximately

• A. $31units$
• B. $51units$
• C. $71units$
• D. $81units$

Answer 1

Answer: (B)

Given, $x=2t^{2}+6t+25 \, and \, y=t^{2}+2t+1 \, $
$\therefore \, \frac{d x}{d t}=4t+6 \, and \, \frac{d y}{d t}=2t+2$
$At \, t=10 \, s$
$\frac{d x}{d t}=4\left(10\right)+6=46 \, and \, \frac{d y}{d t}=2 \, \left(10\right)+2=22$
$v=\sqrt{\left(\frac{d x}{d t}\right)^{2} + \left(\frac{d y}{d t}\right)^{2}}$
$=\sqrt{\left(46\right)^{2} + \left(22\right)^{2}}\simeq 51 \, units$