In fig. 2-27, a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. at time t 0, the red car is at xr 0 and the green car is at xg 220 m. if the red car has a constant velocity of 20 km/h, the cars pass each other at x 44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x 76.6 m. what are (a) the initial velocity and (b) the constant acceleration of the green car?

a) V₀ = 13.5m/s

b) a = 2.1 m/s²

Explanation:

1) Convert velocities to m/s

i) 20 km/h * 1000m/kg * 1h/3600s = 5.556m/s

ii) 40km/h = 11.111m/s

2) Red car, time elapsed to reach position x = 44.5m

Constant velocity ⇒ x = V*t ⇒ t = x / V

⇒ t₁ = 44.5m / 5.556m/s = 8s

3) Red car, time elapsed to reach position x = 77.6m

t₂ = 76.6m / 11.111/s = 6.9s

4) Green car, distance run at t₁ = 8s, x = 44.5m

i) uniform acceleration equation d = V₀t + at² / 2

ii) d = 220m - x = 220m - 44.5m = 175.5m = V₀ (8) + a (8) ² / 2

175.5 = 8V₀ + 32a ↔ equation (1)

5) Green car, distance run at t₂ = 6.9s, x = 76.6m

i) d = 220m - x = 220m - 76.6m = 143.4

ii) 143.4 = V₀t₂ + at₂² / 2

143.4 = V₀ (6.9) + a (6.9) ² / 2

143.4 = 6.9V₀ + 23.8a ↔ equation (2)

6) Solve the system of equations:

175.5 = 8V₀ + 32a ↔ equation (1)

143.4 = 6.9V₀ + 23.8a ↔ equation (2)

V₀ = 13.5m/s

a = 2.1 m/s²

The answer choice is 83