In the figure here, a red car and a green car move toward each other in adjacent

Question

In the figure here, a red car and a green car 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 = 216 m. If the red car has a constant velocity of 20.0 km/h, the cars pass each other at x = 43.6 m. On the other hand, if the red car has a constant velocity of 40.0 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity (in km/h) and (b) the (constant) acceleration (in m/s2) of the green car? Include the signs.

Explanation / Answer

27km/h * 1000m/km * 1h/3600s = 7.5 m/s 54km/h = 15 m/s red car: first case: x(t) = 7.5m/s * t second case: x(t) = 15m/s * t green car: X(t) = 210m - vt - ½at² Find where x = X: First case: t = s / v = 44.5m / 7.5m/s = 5.93s (from red car) Then for green car, 44.5 = 210 - v(5.93) - ½a(5.93)² = 210 - 5.93v - 17.6a and 17.6a = 165.5 - 5.93v and a = 9.4 - 0.337v Second case: t = s / v = 76.4m / 15m/s = 5.09s Then 76.4 = 210 - v(5.09) - ½a(5.09)² = 210 - 5.09v - 12.97a or 0 = 133.6 - 5.09v - 12.97a substitute for a: 5.09v = 133.6 - 12.97(9.4 - 0.337v) = 133.6 - 122 + 4.37v 0.72v = 11.6 v = 16.1 m/s ? green velocity a = 9.4 - 0.337v = 9.4 - 0.337 * 16.1 = 3.97 m/s² ? green acceleration Check: X(5.09s) = 210 - 16.1m/s * 5.09s) - ½ * 3.97m/s² * (5.09s)² = 76.6 m v close enough

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