Question

Amir starts riding his bike up a 200-m-long slope at a speed of 18 km/h, decelerating at 0.20 m/s2 as he goes up. at the same instant, becky starts down from the top at a speed of 6.0 km/h, accelerating at 0.40 m/s2 as she goes down. how far has amir ridden when they pass?

Answer 1

The first thing we must know for this case are the following unit conversions:
 1 km = 1000 m
 1 h = 3600 s
 Now we pass the speeds to m / s:
 v1 = 18 * 1000/3600 = 5m / s
 v2 = 6 * 1000/3600 = 1.66 m / s
 Then, we write the kinematic formulas that describe the problem:
 d = v0 * t + (1/2) * a * t ^ 2
 For Amir,
 d1 = 5 * t - (1/2) * (0.20) * t ^ 2
 For Becky,
 d2 = (1.66) * t + (1/2) * (0.40) * t ^ 2
 On the other hand:
 d1 + d2 = 200
 Adding both equations we have:
 6.66 * t + 0.1 * t ^ 2 = 200
 Rewriting:
 0.1 * t ^ 2 + 6.66 * t - 200 = 0
 Solving the polynomial:
 t = 22.46 s
 Then, for Amir we have:
 d1 = 5 * t - (1/2) * (0.20) * t ^ 2
 d1 = 5 * (22.46) - (1/2) * (0.20) * (22.46) ^ 2
 d1 = 61.85m
 answer:
  Amir has ridden 61.85m when they pass