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 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