Respuesta :
Well, since you noted the direction of travel is “due east”, that clearly means the question is asking about a frame of reference that included planetary rotation into the mix. At the equator, that’s about 460m/s in the eastward direction. Clearly, at the north pole, that’s zero—you’re effectively stationary. So, let’s assume you’re somewhere between the equator and the pole, so we’ll take the average of the two, and say the earth is contributing 230m/sec of eastward velocity. So, at time t=0s, your velocity is 230m/sec; at time t=30s, your velocity is 254m/sec. Thus, you plug those into the formula, and you get (254–230)/(30–0), or 0.8m/sec^2.
Now, if the question was to consider this from a larger frame of reference, we’d also have to take the rotation of the earth around the sun into consideration, which is about 107,000km/hr, or about 29,722m/sec. The problem is that we don’t know if we need to *add* that to the rotational velocity of the earth, and motion of the car, or *subtract* it; that all depends on whether the side of the earth that the car is on is facing the sun, or away from the sun. If we assume that sane people do math experiments on their cars only when the sun is shining, then we need to add the velocity in as well; so we get 29,722+230+24, or (29,976–29,952)(30–0). However, that only works if you do your vehicular calculations in the daytime. If, on the other hand, you’re a dark, brooding vigilante who only comes out after darkness falls to drive around, then we need to adjust our calculations to account for the fact that you’re now going retrograde with respect to the sun. Thus, the calculation would become 29,722–230–24, or (29468–29492)/(30–0), or -0.8m/sec^2.
HOPE THIS HELPS :)
