#### Answer

The least amount of time to travel 5.0 meters without the top block slipping is 1.75 seconds.

#### Work Step by Step

We can find the maximum possible force that the force of static friction can provide for the 3.0-kg block without the 4.0-kg block slipping. Let $m_u$ be the mass of the 4.0-kg block.
$F_f = m_u~g~\mu_s$
$F_f = (4.0~kg)(9.80~m/s^2)(0.60)$
$F_f = 23.52~N$
We can find the acceleration of the bottom block when this force of static friction is provided by the contact between the two blocks. Let $m_b$ be the mass of the 3.0-kg block.
$\sum F = m_b~a$
$F_f - (m_b+m_u)~g~\mu_k = m_b~a$
$a = \frac{F_f - (m_b+m_u)~g~\mu_k}{m_b}$
$a = \frac{(23.52~N) - (3.0~kg+4.0~kg)(9.80~m/s^2)(0.20)}{3.0~kg}$
$a = 3.27~m/s^2$
Since both blocks move together, this will be the acceleration of the system of both blocks. We can find the time it takes to travel 5.0 meters.
$d = \frac{1}{2}at^2$
$t = \sqrt{\frac{2d}{a}}$
$t = \sqrt{\frac{(2)(5.0~m)}{3.27~m/s^2}}$
$t = 1.75~s$
The least amount of time to travel 5.0 meters without the top block slipping is 1.75 seconds.