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Question

The increase in acceleration is $2 m / s^{3}$

Time $t=3 s$

Initial velocity $=0,$ initlal acceleration $=0$

In general, acceleration can be r

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$9 $

Vedclass · Answer

The increase in acceleration is $2 m / s^{3}$

Time $t=3 s$

Initial velocity $=0,$ initlal acceleration $=0$

In general, acceleration can be represented as, $\frac{d a}{d t}=2$

On integration with respect to t and a, for lower limit $t=0$ to $t=t$ and for $a=0$ to $a=a$

$a=2 t$

Acceleration is rate of change of velocity,

$\frac{d v}{d t}=2 t$

$d v=2 t d t$

double integrating with respect to $\mathrm{t}$ and $\mathrm{v}$ for limits, $t=0$ to $t$ and $v=0$ to $v$

$\mathrm{v}=t^{2}$

$\frac{d x}{d t}=t^{2}$

Integrating for limits $t=0$ to $t=3$

$x=(3)^{2}-(0)^{2}=9 m $

A particle moves along a straight line in such a way that it’s acceleration is increasing at the rate of $2 m/s^3$. It’s initial acceleration and velocity were $0,$ the distance covered by it in $t = 3$ second is ........ $m$

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