In the figure here, a red car and a green car move toward each other in adjacent
ID: 2048688 • Letter: I
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 = 212 m. If the red car has a constant velocity of 23.0 km/h, the cars pass each other at x = 43.7 m. On the other hand, if the red car has a constant velocity of 46.0 km/h, they pass each other at x = 76.9 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
When velocity of red car, v = 28 km/hr = 28000/3600 = 7.78 m/s Constant velocity implies acceleration = 0 So, v = distance travelled / time time = distance / v = 43.5 / 7.78 = 5.59 sec this is also the time green car travelled so that they meet each other at x = 43.5 m. let us consider that the direction of red car is positive then the direction of green car will be -ve. Now, x - xo = ut + (1/2) at2 where a = acceleration of green car, t = time both cars moved to meet each other at 43.5 m u = inital velocity of green car. 43.5 - 212 = u * 5.59 + (1/2) a (5.59)2 -168.5 = 5.59 u + 15.62 a ------------------------(I) Now, when velocity of red car, v = 56km/hr = 56000/3600 = 16.11 m/s v = distance / time time = distance / v = 76.2 / 16.11 = 4.73 sec then, x - xo = ut + (1/2) at2 76.2 - 212 = u * 4.73 + (1/2) * a * (4.73)2 -135.8 = 4.73 u + 11.19 a ------------------------(II) Equating eqn (I) and (II) , we get [ -168.5 = 5.59 u + 15.62 a ] * 11.19 implies - 1885.51 = 62.55 u + 174.79 a [ -135.8 = 4.73 u + 11.19 a ] * 15.62 implies - 2121.20 = 73.88 u + 174.79 a hence, 235.69 = - 11.33 u u = - 235.69 / 11.33 = - 20.80 m/s note -ve sign accounts for the opposite dirn relative to red car which we assumed to be positive. substituting u in eqn (II) we get -135.8 = 4.73 * (-20.80) + 11.19 a -135.8 = - 98.38 + 11.19 a a = - 37.42 / 11.19 = - 3.34 m /s2 5 4.21a
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