Question

Need help on B please

Answer 1

The slowdown took 2.05 seconds and the roller coaster made a distance 20.5 during that time.

Step-by-step 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:

`v_2^2 = v_1^2 + 2ad\\F = ma\implies a = (F)/(m)\\v_2^2 = v_1^2 + 2(F)/(m)d`

Let us first determine the distance made:

`v_2^2 = v_1^2 + 2(F)/(m)d\implies\\d = (v_2^2 - v_1^2)/(2F)m = (-320(m^2)/(s^2))/(2\cdot (-7023N) )\cdot 900kg = 20.5 m`

The roller coaster made 20.5 meters during slowing down to 2 m/s. Now calculate the time this took.

`v_2 = at+v_1 = (F)/(m)t + v_1\implies t = (v_2-v_1)/(F)m\\t = (-16(m)/(s))/(-7023N)900kg = 2.05s`

It took 2.05 seconds to achieve the slowdown from 18 to 2 m/s