Kinematics
The inspiration comes while in the trip using subway. Assume that visited stations are evenly spaced in a straight line. The train accelerates for some amount of time, moves with a constant speed for some amount of time, and starts decelerating until it reaches the next station such that the same amount of time spent accelerating and decelerating are the same, and the magnitudes of both are the same. What would be a possible function that illustrates this scenario?A piecewise-constant acceleration is constant over extended time intervals and changing in value discontinuously from one interval to the next. This is physically unrealistic in any real-life situation, since the acceleration would be expected to change more or less smoothly from instant to instant. In realistic models of collisions the acceleration changes smoothly, but it still can be approximated as continuous for simplification.static\css\slides.css /* n-columns layout */ .columns { border: 0px blue solid; flex: 1; padding: 0.2em; } $$ \\[1.5em] $$
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Idul Fitri: 31/3 - 1/4
Weeks 16 Date Feb: 18, 25 Mar: 04, 11, 18, 25 Apr: 01, 08, 15, 22, 29 May: 06, 13, 29, 27 Jun: 03FI5282 Physics in Daily Life (2 credit hours in 16 weeks). Topics Physics and technology, Classical physics concepts in everyday life, Modern physics concepts in everyday life. Outcome Ability to recognize the physical principles used in technology, Ability to explain physics concepts in everyday life phenomena.intro syllabus kinematics equationsThere are three to five kinematics equations reported, where from the first two ($v = v_0 + at$ and $x = x_0 + v_0 t + \tfrac12 at^2$), the third can be obtained, also the fourth and the fifth, by eliminating an unknown. The idea of five equations is that each for one unknown, which are $x - x_0$, $v$ (or $v_t)$, $t$ (or $\Delta t$, $t - t_0$), $v_0$, and $a$.Rearrange of (I5) will produce $$\tag{I6} \begin{array}{rcl} 2a(x - x_0) & = & (v_0 + v)(v - v_0) \\[0.5em] (v + v_0)(v - v_0) & = & 2a(x - x_0) \\[0.5em] v^2 - v_0^2 & = & 2a(x - x_0) \\[0.5em] v^2 & = & v_0^2 + 2a(x - x_0), \end{array} $$ which is the third kinematics equation.From previous equations following can be obtained $$\tag{I4} x = x_0 + v_0 \left( \frac{v - v_0}{a} \right) + \tfrac12 a \left( \frac{v - v_0}{a} \right)^2. $$ It can be simplified into $$\tag{I5} \begin{array}{rcl} 2a(x - x_0) & = & 2v_0 (v - v_0) + (v - v_0)^2 \newline & = & (2v_0 + v - v_0)(v - v_0) \newline & = & (v_0 + v)(v - v_0). \end{array} $$First two equations, with $v = v(t)$ and $x = x(t)$, are $$\tag{I1} v = v_0 + at, $$ $$\tag{I2} x = x_0 + v_0 t + \tfrac12 a t^2. $$ Turn (I1) into following $$\tag{I3} t = \frac{v - v_0}{a}. $$ and then substitute to (I2)With $a(t) = a$, $v(t_0) = v_0$, $x(t_0) = x_0$, $t_0 \ne 0$, the equations are $$\tag{H1} v(t) = v_0 + a(t - t_0), $$ $$\tag{H2} x(t) = x_0 + v_0 (t - t_0) + \tfrac12 a (t - t_0)^2. $$ Set $t_0 = 0$, that makes $v(0) = v_0$, $x(0) = x_0$, will turn them into $$\tag{H3} v(t) = v_0 + at, $$ $$\tag{H4} x(t) = x_0 + v_0 t + \tfrac12 a t^2.With initial condition $v(t_0) = v_0$, velocity becomes $$\tag{G1} v(t) = v_0 + a(t - t_0). $$ With initial condition $x(t_0) = x_0$, position becomes $$\tag{G2} x(t) = x_0 + v_0 (t - t_0) + \tfrac12 a (t - t_0)^2. $$ Are they the same as the first two of common kinematic equations?Integrate velocity $v$ to obtain position $x$ $$\tag{F1} x(t) - x(t_0) = \int_{t_0}^t \left[ v_0 + \int_{t_0}^{\tau_2} a(\tau_1) d\tau_1 \right] d\tau_2. $$ Initial condition $x(t_0) = x_0$, will give $$\tag{F2} x(t) = x_0 + v_0 (t - t_0) + \int_{t_0}^t \int_{t_0}^{\tau_2} a(\tau_1) d\tau_1 d\tau_2. $$Integrate acceleration $a$ to obtain velocity $v$ $$\tag{E1} v(t) - v(t_0) = \int_{t_0}^t a(\tau_1) d\tau_1. $$ Initial condition $v(t_0) = v_0$, will give $$\tag{E2} v(t) = v_0 + \int_{t_0}^t a(\tau_1) d\tau_1. $$Integrate acceleration $a$ twice to obtain velocity $v$ and position $x$ $$\tag{D1} v(t) - v(t_0) = \int_{t_0}^t a(\tau) d\tau, $$ $$\tag{D2} x(t) - x(t_0) = \int_{t_0}^t v(\tau) d\tau. $$ Initial conditions are $v(t_0) = v_0$, $x(t_0) = x_0$, and $t_0 = 0$.