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

Answer 2

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

$-x= 5t - 0.2t^2\\\\ x = -5t + 0.2t^2 \ ---- (1)$

Distance traveled by Becky

6 km/h = 1.67 m/s

$200+ x = 1.67t + 0.4t^2\\\\ x = 1.67t +0.4t^2 - 200 \ \ ----(2)$

Solve equation 1 and 2 together

$1.67t + 0.4t^2 - 200 = -5t + 0.2t^2\\\\0.2t^2 + 6.67t -200 = 0\\\\ t =19.1 \ s$

The distance traveled by Amir is calculated as follows;

$x = -5t + 0.2t^2\\\\x = -5(19.1) + 0.2(19.1)^2\\\\x = -22.54 \ m\\\\|x| = 22.54 \ m$

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

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