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

- 8 m / s^{2}

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5 m / s^{2}

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0 m / s^{2}

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- 4 m / s^{2}

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Solution

The correct option is D $- 4 m / s^{2}$
AS Given,
$x = 5 t - 2 t^{2}$
$y = 10 t$

So, $\frac{d x}{d t} = 5 - 4 t$ and $\frac{d^{2} x}{d t^{2}} = - 4 \Rightarrow a_{x}$

Similarly, $a_{y} = \frac{d^{2} y}{d t^{2}} = 0$
At $t = 2 s,$
$\to a =_{x}^i +_{y}^J$
$a_{x} = - 4 m / s^{2} a n d a_{y} = 0 m / s^{2}$
$\to a = - 4 m / s^{2}$

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