The relation between the displacement x (in m | vedclass

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Question

(B) Given the relation: $t = 2x^2 + 3x$. To find velocity $v$,differentiate $t$ with respect to $x$: $\frac{dt}{dx} = 4x + 3$. Since $v = \frac{dx}{dt

Vedclass · Accepted Answer

$-\frac{1}{16} \ ms^{-2}$

Vedclass · Answer

(B) Given the relation: $t = 2x^2 + 3x$. To find velocity $v$,differentiate $t$ with respect to $x$: $\frac{dt}{dx} = 4x + 3$. Since $v = \frac{dx}{dt}$,we have $v = \frac{1}{4x + 3} = (4x + 3)^{-1}$. To find acceleration $a$,differentiate $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v$. $\frac{dv}{dx} = -1(4x + 3)^{-2} \cdot 4 = -\frac{4}{(4x + 3)^2}$. Thus,$a = -\frac{4}{(4x + 3)^2} \cdot \frac{1}{4x + 3} = -\frac{4}{(4x + 3)^3}$. Given displacement $x = 25 \ cm = 0.25 \ m = \frac{1}{4} \ m$. Substitute $x = \frac{1}{4}$ into the acceleration formula: $a = -\frac{4}{(4(1/4) + 3)^3} = -\frac{4}{(1 + 3)^3} = -\frac{4}{4^3} = -\frac{4}{64} = -\frac{1}{16} \ ms^{-2}$.

The relation between the displacement $x$ (in metre) and the time $t$ (in second) of a particle is $t = 2x^2 + 3x$. If the displacement of the particle is $25 \ cm$ from the origin $(x = 0)$,then the acceleration of the particle is:

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