A body of mass 1 , kg starts moving from rest | vedclass

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(B) Given $K.E. = \frac{1}{2}mv^2 = 2t^2$. Since $m = 1 \, kg$,we have $\frac{1}{2}(1)v^2 = 2t^2$,which implies $v^2 = 4t^2$,so $v = 2t \, m/s$.Tangen

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power of all forces at $t = 2 \, s$ is $8 \, W$

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(B) Given $K.E. = \frac{1}{2}mv^2 = 2t^2$. Since $m = 1 \, kg$,we have $\frac{1}{2}(1)v^2 = 2t^2$,which implies $v^2 = 4t^2$,so $v = 2t \, m/s$.Tangential acceleration $a_t = \frac{dv}{dt} = \frac{d}{dt}(2t) = 2 \, m/s^2$. Thus,option $A$ is incorrect.Power $P = \frac{d(K.E.)}{dt} = \frac{d}{dt}(2t^2) = 4t$. At $t = 2 \, s$,$P = 4(2) = 8 \, W$. Thus,option $B$ is correct.Tangential force $F_t = m \cdot a_t = 1 \cdot 2 = 2 \, N$. Thus,option $D$ is incorrect.For the first round,distance $s = 2\pi r = 2\pi(8) = 16\pi \, m$. Since $v = 2t$,$s = \int v \, dt = \int_0^T 2t \, dt = T^2$. Setting $T^2 = 16\pi$,$T = 4\sqrt{\pi} \, s 
eq 2 \, s$. Thus,option $C$ is incorrect.

$A$ body of mass $1 \, kg$ starts moving from rest at $t = 0$ in a circular path of radius $8 \, m$. Its kinetic energy varies as a function of time as: $K.E. = 2t^2 \, J$,where $t$ is in seconds. Then:

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