The block will slide once the deceleration of the truck exceeds μg.
Once the block begins to slide, its absolute deceleration is constant at μg.
So the stopping distance of the block is v^2 / (2μg) = 40.77 m.
And the block is allowed to stop within 3 m more than the truck. So the minimum stopping distance of the truck is 37.77 m.
If you use g = 9.81, you get 37.77 m
If g = 10, then you get 37 m.
but his answer is a bit short, and hard to understand...
but his answer is a bit short, and hard to understand...
Rather than hunting for answers, your time might be better spent trying to figure it out on your own. Use the hints I gave and your knowledge of Newton's laws.