Question

rw n8.00-09:15 - HY 2053 Test #1 2019-01-30 1. A subway train starts from rest at a stat ion and accelerates at a constant acceleration of a 2.85 m/2 for 22.3 s. It then travels at a constant speed for 62.0 s and th at a constant acceleration o f a2 = 3.75 A until it stops at the next station. Find the total distance covered by the subway train for this trip

Answer 1

This question is based on the following concept

1. Newton's equation of motion

2. Constant speed motion also called uniform motion

3. Decelerating motion

The detailed solution is described below

Newton's equation of motion Constant speed motion also called uniform motion Decelerating motion The detailed solution is described below Let us go the basics first. U = 0 (rest) a_1 = 2.85 m/s^2 t_1 = 22.3 s From, Newton's eqn of motion distance s_1 = ut_1 + 1/2 a_1 t_1^2 s_1 = 0 + 1/2 * 2.85 * 22.3^2 s_1 = 708.638 m Now, v_1 = u + a_1 t_1 v_1 = 0 + 2.85 * 22.3 v_1 = 63.555 m/s Now, for second phase of motion v = constant = v_1 t_2 = 62 s Thus, distance s_2 = v_1 t_2 = 63.555 * 62 s_2 = 3940.41 m \r\n Now, for last phase of motion a_2 = -3.75 m/s^2 v_2 = 0 (stops) u_2 = v_1 = 63.555 m/s Now, again from Newton's eqn: - v_2^2 = u_2^2 + 2a_2 s_3 (III eqn of motion) 0^2 = 63.555^2 + 2 * (-3.75) * s_3 s_3 = 538.565 m