A red car and a green car, identical except for the color, move toward each othe

Question

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 = 43.9 m, and if it has a constant velocity of +40 km/h, they pass each other at x = 77.1 m.

(a) What is the initial velocity of the green car? (Indicate direction with the sign of your answer.)
? m/s

(b) What is the acceleration of the green car? (Indicate direction with the sign of your answer.)
? m/s2

Explanation / Answer

one of the possible reasons you did wrong is you forgot to change to a same unit. 20 km/hr = 50/9 m/s find the time when they pass each other 43.5 / (50/9) = 7.83 secs red car = (50/9) (7.83) green car = 0.5(a)(7.83)² + 7.83V + 220 when the two car meets, they are in the same position (50/9) (7.83) = 0.5(a)(7.83)² + 7.83V + 220 given that if the red car travels at 40 km/hr (100/9 m/s), they'd pass each other at 77.3m same process. Find the time when they pass each other t = 77.3 / (100/9) = 6.957s red car = (100/9) (6.957) green car = 0.5(a)(6.957)² + 6.957V + 220 (100/9) (6.957) = 0.5(a)(6.957)² + 6.957V + 220 ---- so you have two equations: (50/9) (7.83) = 0.5(a)(7.83)² + 7.83V + 220 (100/9) (6.957) = 0.5(a)(6.957)² + 6.957V + 220 solve for a and V and you'll get: a = -4.650154 m/s² and V = -4.336154 m/s or you can convert to km/hr² and km/hr a = -60,265.9958 km / hr^2 V = -15.6101544 km / hr the acceleration and velocity are negative because the green car travels left or opposite of the red car

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