The $x$ and $y$ coordinates of the particle at any time are $x = 5 t - 2 t^{2} a n d y = 10 t$ respectively, where $x$ and $y$ are in meters and $t$ in seconds. The acceleration of the particle at $t = 2 s e c$ is:

$0 m / s e c^{2}$

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$5 m / s e c^{2}$

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$- 4 m / s e c^{2}$

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$8 m / s e c^{2}$

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Solution

The correct option is C
$- 4 m / s e c^{2}$

Given
$x = 5 t - 2 t^{2}$
$y = 10 t$
On differentiating, we get,
$d x / d t = 5 - 4 t$
$d y / d t = 10$
$a_{x} = \frac{d^{2} x}{d t^{2}} = - 4 . . . . . . . . . . . . . . . . . . . . . . (1) a_{y} = \frac{d^{2} y}{d t^{2}} = 0 . . . . . . . . . . . . . . . . . . . . . . . . . (2)$
Adding equations $(1)$ and $(2)$ , we get,
$a = a_{x} + a_{y}$
$a = - 4 + 0$
$a = - 4$
Hence, the acceleration of the particle at $t = 2 s$ is $- 4 m / s^{2}$

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