Answer:
4m (with the assumption that the deceleration is same in both cases)
Explanation:
The third equation of a uniformly decelerated motion is used in solving this problem.
[tex]v^2=u^2-2as.................(1)[/tex]
where v is the final velocity, u the initial velocity, s the distance covered and a the deceleration.
The negative sign is due to the fact that the body under consideration is decelerating.
Given;
u = 0.5m/s
v = 0m/s (because it comes to rest)
s = 1m
a = ?
Substituting all values into equation (1);
[tex]0^2=0.5^2-2*a*1\\2a=0.5^2\\2a=0.25\\a=\frac{0.25}{2}\\a=0.125m/s^2[/tex]
Assuming the block maintained this same deceleration, but with an initial velocity of 1m/s, then the distance travelled would be as follows;
u = 1m/s
v  = 0m/s
s = ?
a = [tex]0.125m/s^2[/tex]
By still using equation (1), we obtain the following;
[tex]0^2=1^2-2*0.125*s\\0.25s=1\\s=\frac{1}{0.25}\\s=4m[/tex]
