A particle starts from the origin at time t=0 | vedclass

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(A) Given that the velocity $v$ is a function of time: $v(t) = 10t 	ext{ cm/s}$.Since the particle starts from the origin at $t=0$, the initial veloc

Vedclass · Accepted Answer

$320$

Vedclass · Answer

(A) Given that the velocity $v$ is a function of time: $v(t) = 10t 	ext{ cm/s}$.Since the particle starts from the origin at $t=0$, the initial velocity $u = v(0) = 0 	ext{ cm/s}$.The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(10t) = 10 	ext{ cm/s}^2$.Using the second equation of motion for constant acceleration, the distance $s$ covered in time $t$ is given by:$s = ut + \frac{1}{2}at^2$Substituting the values $u = 0 	ext{ cm/s}$, $a = 10 	ext{ cm/s}^2$, and $t = 8 	ext{ s}$:$s = (0)(8) + \frac{1}{2} 	imes 10 	imes (8)^2$$s = 0 + 5 	imes 64$$s = 320 	ext{ cm}$.Thus, the distance covered by the particle in $8 	ext{ s}$ is $320 	ext{ cm}$.

$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)

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