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Byju's Answer

Standard XII

Physics

Acceleration for Non Linear Motion

The x and ...

Question

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:

0

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5 m / s^{2}

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- 4 m / s^{2}

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- 8 m / s^{2}

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Solution

The correct option is C $- 4 m / s^{2}$
$x = 5 t - 2 t^{2}$
$y = 10 t$
$\frac{d x}{d t} = 5 - 4 t$
$a_{x} = \frac{d^{2} x}{d t^{2}} = - 4$
$a_{y} = \frac{d^{2} y}{d t^{2}} = 0$
$a = a_{x} + a_{y} = - 4 + 0$
$a = - 4 m s^{2}$

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Similar questions

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

Q. 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

m e t e r s

and

t

s e c o n d s .

The acceleration of the particle at

t = 2 s

Q. The x and y coordinates of a particle at any time

t

are given by

x = 2 t + 4 t^{2}

and

y = 5 t

, where x and y are in meter and

t

in second. The acceleration of the particle at

t = 5 s

is:

Q. The

x

and

y

coordinate of a particle at any time t are given by

x = 7 t + 4 t^{2}

and

y = 5 t

, where

x

and

y

are in meter and

t

in seconds. The acceleration of particle at

t = 5 s

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The x and y coordinates of the particle at any time are x=5t−2t2 and y=10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t=2 s is:

The correct option is C −4 m/s2x=5t−2t2y=10tdxdt=5−4tax=d2xdt2=−4ay=d2ydt2=0a=ax+ay=−4+0a=−4 ms2

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:

The correct option is C $- 4 m / s^{2}$
$x = 5 t - 2 t^{2}$
$y = 10 t$
$\frac{d x}{d t} = 5 - 4 t$
$a_{x} = \frac{d^{2} x}{d t^{2}} = - 4$
$a_{y} = \frac{d^{2} y}{d t^{2}} = 0$
$a = a_{x} + a_{y} = - 4 + 0$
$a = - 4 m s^{2}$