The x and y coordinates of the particle at any time are - -2-2 and - - 10- respectively- where - and - are in meters and t in seconds- The acceleration of the particle at - - 2 sec is in m-s2

Given: x=2t^2, y=10t We know that, a_x=d^2xdt^2=d(d(2t^2)dt)dt=d(4t)dt=4 Similarly, a_y=d^2ydt^2=d(d(10t)dt)dt=d(10)dt=0 So, we have a=a_xî+a_yĵ=4î+0ĵ=4î Thus, ...

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Standard XIII

Physics

Distance and Displacement

The x and y c...

Question

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

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Solution

Given:
$x = 2 t^{2}, y = 10 t$
We know that,
$a_{x} = \frac{d^{2} x}{d t^{2}} = \frac{d (\frac{d (2 t^{2})}{d t})}{d t} = \frac{d (4 t)}{d t} = 4$
Similarly, $a_{y} = \frac{d^{2} y}{d t^{2}} = \frac{d (\frac{d (10 t)}{d t})}{d t} = \frac{d (10)}{d t} = 0$
So, we have
$\to a = a_{x}^i + a_{y}^j = 4^i + 0^j = 4^i$
Thus, $a = 4 {m/s}^{2}$ along positive $x -$ axis
Hence, option $(A)$ is correct.

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The x and y coordinates of the particle at any time are x=2t2 and y=10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t=2 sec is (in m/s2)

Given: x=2t2,y=10t We know that, ax=d2xdt2=d(d(2t2)dt)dt=d(4t)dt=4 Similarly, ay=d2ydt2=d(d(10t)dt)dt=d(10)dt=0 So, we have →a=ax^i+ay^j=4^i+0^j=4^i Thus, a=4 m/s2 along positive x−axis Hence, option (A) is correct.

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