Question

The position of a particle moving along x axis is given by x =(-2t^3 + 3t^2 + 5) m. The acceleration of a particle at instant its velocity become zero is

Answer 1

Please find below the answer

Answer 2

Given distance $x=-2{ t }^{ 3 }+3{ t }^{ 2 }+5$

Since we know velocity is distance over time so $v=\frac { dx }{ dt }$

Differentiating gives $v=-2x(3{ t }^{ 2 })+3x(2t)\quad \Rightarrow v=-6{ t }^{ 2 }+6t$

The instance $v=0 \Rightarrow\quad 0= -6{ t }^{ 2 }+6t \Rightarrow\quad t(-6t+6)= 0$

$\Rightarrow$        This expression gives t=0 or t=1

Now acceleration is velocity by time $\Rightarrow a=\frac { dv }{ dt } =-6x(2t)+6$

So when  t=0 we get  a=6 while when t=1 $\Rightarrow a=-6$