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Standard XII

Mathematics

Average Rate of Change

The motion of...

Question

The motion of a particle along a straight line described by the function $x = (2 t - 3)^{2}$ , when is in metre and $t$ is in second. Then, the velocity of the particle at origin is

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Solution

The correct option is A $0$
Given that, $x = (2 t - 3)^{2}$
On differentiating w.r.t. $x$ , we get
$\frac{d x}{d t} = 2 (2 t - 3) (2)$
At origin, $x = 0$
$∴ 2 t - 3 = 0 \Rightarrow t = \frac{3}{2}$
Therefore, $Velocity = \frac{d y}{d x} = 4 (2 \times \frac{3}{2} - 3) = 0$ .

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The motion of a particle along a straight line described by the function x=(2t−3)2, when is in metre and t is in second. Then, the velocity of the particle at origin is

The correct option is A 0Given that, x=(2t−3)2On differentiating w.r.t. x, we getdxdt=2(2t−3)(2)At origin, x=0∴2t−3=0⇒t=32Therefore, Velocity=dydx=4(2×32−3)=0.

The motion of a particle along a straight line described by the function $x = (2 t - 3)^{2}$ , when is in metre and $t$ is in second. Then, the velocity of the particle at origin is

The correct option is A $0$
Given that, $x = (2 t - 3)^{2}$
On differentiating w.r.t. $x$ , we get
$\frac{d x}{d t} = 2 (2 t - 3) (2)$
At origin, $x = 0$
$∴ 2 t - 3 = 0 \Rightarrow t = \frac{3}{2}$
Therefore, $Velocity = \frac{d y}{d x} = 4 (2 \times \frac{3}{2} - 3) = 0$ .