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(B) Given that the rate of change of acceleration is constant: $\frac{da}{dt} = 2  m/s^3$. Integrating with respect to time $t$ with initial condition

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Given that the rate of change of acceleration is constant: $\frac{da}{dt} = 2  m/s^3$. Integrating with respect to time $t$ with initial conditions $a(0) = 0$: $a = \int 2  dt = 2t$. Since acceleration $a = \frac{dv}{dt}$,we have $\frac{dv}{dt} = 2t$. Integrating with respect to time $t$ with initial velocity $v(0) = 0$: $v = \int 2t  dt = t^2$. Since velocity $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = t^2$. Integrating to find the distance $x$ covered from $t = 0$ to $t = 3  s$: $x = \int_{0}^{3} t^2  dt = \left[ \frac{t^3}{3} \right]_{0}^{3} = \frac{3^3}{3} - 0 = \frac{27}{3} = 9  m$.

$A$ particle moves along a straight line in such a way that its acceleration is increasing at the rate of $2 m/s^3$. Its initial acceleration and velocity were $0$. The distance covered by it in $t = 3 s$ is ........ $m$.

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