Question

A particle moves such that its x and y coordinates as a function of time are x = 2+ $5t^2$ and y = $t^3$. Acceleration of particle at t = 2 sec is

• A.
  12 m/$s^2$
• B.
  $\sqrt{244}$ m/$s^2$
• C.
  $\sqrt{136}$ m/$s^2$
• D.
  0 m/$s^2$

Answer 1

The correct option is B

$\sqrt{244}$ m/$s^2$
$\overrightarrow{v}$ = $\frac{d}{dt}$(x$\hat{i}$ + y$\hat{j}$) = 10t$\hat{i}$ + 3$t^2$$\hat{j}$
$\overrightarrow{a}$ = $\frac{d\left(\overrightarrow{v}\right)}{dt}$ = 10$\hat{i}$ + 6t$\hat{j}$
$\Rightarrow$ at t = 2 sec
$\overrightarrow{a}$ = 10$\hat{i}$ + 12$\hat{j}$
$\Rightarrow$ I$\overrightarrow{a}$I = $\sqrt{{10}^2+{12}^2}$
$\Rightarrow$ $\sqrt{100+144}$
$\Rightarrow$ $\sqrt{244}$ m/$s^2$