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(B) The velocity of the particle is given by $v = 4t$. Since the particle starts from rest,the initial velocity $u = 0$.We know that the distance $S$

Vedclass · Accepted Answer

$32$

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(B) The velocity of the particle is given by $v = 4t$. Since the particle starts from rest,the initial velocity $u = 0$.We know that the distance $S$ covered by a particle is given by the integral of velocity with respect to time:$S = \int_{0}^{t} v \ dt$Substituting the given expression for $v$:$S = \int_{0}^{4} 4t \ dt$$S = 4 \left[ \frac{t^2}{2} ight]_{0}^{4}$$S = 2 	imes [t^2]_{0}^{4}$$S = 2 	imes (4^2 - 0^2)$$S = 2 	imes 16 = 32 \ m$Alternatively,using the kinematic equation $v = u + at$,we compare $v = 4t$ with $v = 0 + at$ to find acceleration $a = 4 \ m s^{-2}$.Using $S = ut + \frac{1}{2}at^2$:$S = 0(4) + \frac{1}{2}(4)(4)^2 = 2 	imes 16 = 32 \ m$.

Column-$I$	Column-$II$
$(1)$ Velocity-time relation	$(a)$ $v = v_0 + at$
$(2)$ Velocity-displacement relation	$(b)$ $S = v_0t + \frac{1}{2}at^2$
	$(c)$ $v^2 = v_0^2 + 2as$

The velocity $(v)$ of a particle starting from rest increases linearly with time $(t)$ as $v = 4t$,where $v$ is in $m s^{-1}$ and $t$ is in seconds. The distance covered by the particle in the first $4$ seconds is (in $m$)

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Column-$I$ Column-$II$

$(1)$ Velocity-time relation $(a)$ $v = v_0 + at$

$(2)$ Velocity-displacement relation $(b)$ $S = v_0t + \frac{1}{2}at^2$

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