Answer:
The slowdown took 2.05 seconds and the roller coaster made a distance 20.5 during that time.
Explanation:
Start with the kinematic equation for velocity. Let v1 and v2 be the initial and the new velocity, respectively. Let s denote the distance made duringthe slowdown. Let m denote the mass of the roller coaster and "a" its acceleration (negative = deceleration). the equation is as follows:
[tex]v_2^2 = v_1^2 + 2ad\\F = ma\implies a = \frac{F}{m}\\v_2^2 = v_1^2 + 2\frac{F}{m}d[/tex]
Let us first determine the distance made:
[tex]v_2^2 = v_1^2 + 2\frac{F}{m}d\implies\\d = \frac{v_2^2 - v_1^2}{2F}m = \frac{-320\frac{m^2}{s^2}}{2\cdot (-7023N) }\cdot 900kg = 20.5 m[/tex]
The roller coaster made 20.5 meters during slowing down to 2 m/s. Now calculate the time this took.
[tex]v_2 = at+v_1 = \frac{F}{m}t + v_1\implies t = \frac{v_2-v_1}{F}m\\t = \frac{-16\frac{m}{s}}{-7023N}900kg = 2.05s[/tex]
It took 2.05 seconds to achieve the slowdown from 18 to 2 m/s
