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

Question

Dr. Abubacker · Accepted Answer

(b) Given, $x = 2t^{2} + 6t + 25\ and\ y = t^{2} + 2t + 1\ $
$	herefore\frac{dx}{dt} = 4t + 6\ and\frac{dy}{dx} = 2 + 2$
$At\ t = 10\ s$
$\frac{dx}{dt} = 4(10) + 6 = 46\ and\ \frac{dy}{dt} = 2\ (10) + 2 = 22$
$v = \sqrt{\left( \frac{dx}{dt} ight)^{2} + \left( \frac{dy}{dt} ight)^{2}}$
$= \sqrt{(46)^{2} + (22)^{2}}\  \simeq 51m/s$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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