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?

The distance traveled by Amir when they passed is 22.54 m.

Time of motion

The distance traveled by Amir and Becky before they meet will occur at a particular time t.

Distance traveled by Amir

18 km/h --------> 5 m/s

Distance traveled by Becky

6 km/h = 1.67 m/s

Solve equation 1 and 2 together

The distance traveled by Amir is calculated as follows;

Thus, the distance traveled by Amir when they passed is 22.54 m.

Learn more about time of motion here: link

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
 
  Amir has ridden 61.85m when they pass

Where are the options ?

What even is this. where is the question