Two bodies of masses m1 and m2 are acted upon by equal forces F for equal time t. If the ratio of their masses is m1:m2 = 4:3, find the ratio of distances covered by them.

Given that the force $F$ is equal for both bodies, and their masses are $m_1$ and $m_2$.
According to Newton's Second Law, $F = ma$.
For body 1: $F = m_1 a_1 \implies a_1 = \frac{F}{m_1}$
For body 2: $F = m_2 a_2 \implies a_2 = \frac{F}{m_2}$

Both bodies start from rest ($u=0$) and act for equal time $t$.
The equation for distance covered is $s = ut + \frac{1}{2}at^2$.
Since $u=0$, the equation simplifies to $s = \frac{1}{2}at^2$.
For body 1: $s_1 = \frac{1}{2} a_1 t^2 = \frac{1}{2} \left(\frac{F}{m_1}\right) t^2$
For body 2: $s_2 = \frac{1}{2} a_2 t^2 = \frac{1}{2} \left(\frac{F}{m_2}\right) t^2$

Given the ratio of their masses $m_1:m_2 = 4:3$. This means $\frac{m_1}{m_2} = \frac{4}{3}$.
Therefore, $\frac{m_2}{m_1} = \frac{3}{4}$.
So, the ratio of their distance covered is $s_1:s_2 = 3:4$.

The correct option is (a).

The displacement $s$ is given by $s = \frac{1}{3}t^3 + \frac{1}{2}t^2 - 12t + 5$.
Velocity $v$ is the derivative of displacement with respect to time $t$:

When the butterfly is momentarily at rest, its velocity $v=0$.

We can factor the quadratic equation:

This gives two possible values for $t$:

Since time cannot be negative, we take the positive value.

The butterfly is momentarily at rest at $t = 3 \text{ s}$.

The correct option is (b).