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 distances covered by a particle thrown in a vertical plane, in horizontal and vertical directions at any instant of time

t

are

x = 3 t

and

y = 4 t - 4 t^{2}

. The range of the particle 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 coordinates of a particle moving in x-y plane at any instant of time t are x=4t2;y=3t2. The speed of the particle at that instant is :

The correct option is A 10tx=4t2Differentiate with respect to t, we getVx=dxdt=2×4t=8tSimilarly, Vy=dydt=2×3t=6tThe speed of the particleV=√V2x+V2y=√(8t)2+(6t)2=√64t2+36t2√100t2=10t

The coordinates of a particle moving in x-y plane at any instant of time t are $x = 4 t^{2}; y = 3 t^{2}$ . The speed of the particle at that instant is :