A block of mass 1 kg starts from rest at x=0 | vedclass

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(A) Given: Mass $m = 1 \ kg$, Force $F = kt$, where $k = 1 \ Ns^{-1}$.According to Newton's second law, $F = ma$.Since $a = \frac{dv}{dt}$, we have $m

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$36$

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(A) Given: Mass $m = 1 \ kg$, Force $F = kt$, where $k = 1 \ Ns^{-1}$.According to Newton's second law, $F = ma$.Since $a = \frac{dv}{dt}$, we have $m \frac{dv}{dt} = kt$.Substituting $m = 1$ and $k = 1$, we get $\frac{dv}{dt} = t$.Integrating with respect to time $t$ (starting from rest, $v=0$ at $t=0$):$v = \int t \ dt = \frac{t^2}{2}$.Since $v = \frac{dx}{dt}$, we have $\frac{dx}{dt} = \frac{t^2}{2}$.Integrating again to find the displacement $x$ at $t = 6 \ s$:$x = \int_{0}^{6} \frac{t^2}{2} \ dt = \left[ \frac{t^3}{6} \right]_{0}^{6}$.$x = \frac{6^3}{6} = \frac{216}{6} = 36 \ m$.

$A$ block of mass $1 \ kg$ starts from rest at $x=0$ and moves along the $X$-axis under the action of a force $F=kt$, where $t$ is time and $k=1 \ Ns^{-1}$. The distance the block will travel in $6 \ seconds$ is: (in $m$)

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