Two bodies, A and B, are moving with uniform velocities. The ratio...

Let the masses of bodies A and B be $m_{\text{A}}$ and $m_{\text{B}}$, and their linear momenta be $p_{\text{A}}$ and $p_{\text{B}}$.

Given:
$\dfrac{m_{\text{A}}}{m_{\text{B}}} = \dfrac{1}{4}$
$\dfrac{p_{\text{A}}}{p_{\text{B}}} = \dfrac{2}{1}$

The velocity of a body is given by $v = \dfrac{p}{m}$.
The distance covered in time $t$ is $d = v \times t = \dfrac{p}{m} \times t$.

Since the time interval $t$ is the same for both bodies, the ratio of distances covered is:
$\dfrac{d_{\text{A}}}{d_{\text{B}}} = \dfrac{p_{\text{A}} / m_{\text{A}}}{p_{\text{B}} / m_{\text{B}}} = \left( \dfrac{p_{\text{A}}}{p_{\text{B}}} \right) \times \left( \dfrac{m_{\text{B}}}{m_{\text{A}}} \right)$

Substituting the given values:
$\dfrac{d_{\text{A}}}{d_{\text{B}}} = \left( \dfrac{2}{1} \right) \times \left( \dfrac{4}{1} \right) = \dfrac{8}{1}$

Hence, the ratio of the distances covered is $8 : 1$.

Answer: $8 : 1$