A train starts from rest and accelerates uniformly,until it has traveled 5.6 km and acquired a velocity of 42 m/s.The train then moves at a constant velocity of 42 m/s for 420 s.The train then slows down uniformly at 0.065 m/s²,until it is brought to a halt.The acceleration during the first 5.6 km of travel is closest to: A)0.16 m/s² B)0.14 m/s² C)0.17 m/s² D)0.19 m/s² E)0.20 m/s²

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Explanation:The acceleration during the first 5.6 km of travel can be found using the formula:a = (v^2 - u^2) / 2swhere:a is the accelerationv is the final velocityu is the initial velocitys is the distance traveledIn this case, the initial velocity is 0 m/s, the final velocity is 42 m/s, and the distance traveled is 5.6 km = 5600 m. Substituting these values into the formula, we get:a = (42^2 - 0^2) / 2(5600)a = 1764 / 11200a = 0.1575 m/s^2Therefore, the acceleration during the first 5.6 km of travel is closest to 0.16 m/s^2.

A Train Starts from Rest and Accelerates Uniformly,until It Has