Answer:
[tex]60\; {\rm ft \cdot s^{-1}}[/tex].
Explanation:
Let [tex]a[/tex] denote the acceleration of this car; [tex]a = (-18)\; {\rm ft \cdot s^{-2}}[/tex]. Note that acceleration [tex]a\![/tex] measures the rate of change in velocity. Since this car is slowing down, acceleration [tex]a\!\![/tex] will be negative.
Let [tex]x[/tex] denote the displacement of the car. It is given that [tex]x = 100\; {\rm ft}[/tex] as the car braked.
Let [tex]v[/tex] denote the velocity of the car after braking. Since the car has stopped, [tex]v = 0\; {\rm ft \cdot s^{-1}}[/tex].
Let [tex]u[/tex] denote the initial velocity of the car. Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find the value of [tex]u\![/tex]. Rearrange this equation to obtain:
[tex]u^{2} = v^{2} - 2\, a\, x[/tex].
[tex]\displaystyle u = \sqrt{v^{2} - 2\, a\, x}[/tex].
Substitute in [tex]v = 0\; {\rm ft \cdot s^{-1}}[/tex], [tex]a = 18\; {\rm ft \cdot s^{-2}}[/tex], and [tex]x = 100\; {\rm ft}[/tex] to find the value of [tex]u[/tex]:
[tex]\begin{aligned} u &= \sqrt{v^{2} - 2\, a\, x} \\ &= \left(\sqrt{0^{2} - 2 \times 18 \times 100}\right)\; {\rm ft \cdot s^{-1}} \\ &= 60\; {\rm ft \cdot s^{-1}}\end{aligned}[/tex].
In other words, this car was travelling at [tex]60\; {\rm ft \cdot s^{-1}}[/tex] before braking.