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 (x)r = 0 and the green car is at (x)g = 223 m. If the red car has a constant velocity of 29.0 km/h, the cars pass each other at x = 43.2 m. On the other hand, if the red car has a constant velocity of 58.0 km/h, they pass each other at x = 76.1 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.

Tried the problem several times over and still get it wrong ;/ and help would be appreciated

Explanation / Answer

a)
given that red car velocity is 29 km/h and it passes green car at 43.2m
time taken for both cars to meet = distance / speed = (0.0432 - 0)/29 = 1.49E-3 hr
speed of green car = distance / time = (0.223 - 0.0432) / 1.49E-3 = 120.76 km/hr

given that red car velocity is 58 km/h and it passes green car at 76.1m
time taken for both cars to meet = distance / speed = (0.0761 - 0)/58 = 1.31E-3 hr
speed of green car = distance / time = (0.223 - 0.0761) / 1.31E-3 = 112.14 km/hr

t = 1.49E-3 hr , v = 120.76 km/h
t = 1.31E-3 hr , v = 112.14 km/h

solve it as point of geometery, (1.49E-3 , 120.76) and (1.31E-3 , 112.14)
gradient = 4.79E4

v = 4.79E4t + c
120.76 = 4.79E4(1.49E-3) + c
c = 49.39

v = 4.79E4t + 49.39
when t is 0, v is 49.39

given that t = 0, green car is travelling at 49.39 km/h

b) from the eqn below, v is the velocity and t is the time.

v = 4.79E4t + 49.39

acceleration is dv/dt
therefore, acceleration is 4.79E4 = 47900 m/s2

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