Answer:
It would slide a distance 4d
Explanation:
From the first scenario where the speed is v and the distance sliding is d we can calculate de value of the acceleration:
[tex]V_{f}^{2} =V_{o}^{2} - 2.a.x[/tex]
We know that
[tex]V_{f}=0[/tex]
[tex]V_{o}=v[/tex]
[tex]x=d[/tex]
So, the acceleration is [tex]a=\frac{v^{2}}{2d}[/tex]
Now, using the same formula and with the new value for [tex]V_{o}=2v[/tex] we can solve for x from:
[tex]0=(2v)^{2} -2.\frac{v^{2}}{2d} .x[/tex]
x=4d