Mechanics Homework Motion in One Dimenston | University of New Haven

University:
University of New Haven
Course:
PHYS 1150 | Mechanics, Heat, And Waves With Laboratory
Academic year:

2023
Views:

477

Pages:

4
Author:

Baron Vaughan

HAPTER TWO MOTION IN ONE DIMENSION (along a straight line) x depends on time (t), which means it is a function of time Vertical direction is the same as horizontal (just turned 90°) At t=t1 [X=X1\] At t=t2 \[x=x2\] Δx=x2−x1 Displacement [m][m/s] Vx, AVG = (X2 - X1) / (t2 - t1) AVERAGE VELOCITY Secant line limit t2 -> t1 (Δt=0) Ax (rise ti + Δε tz t (run) Slope tangent line d=very small dx Slope lim. (derivative) 2 at Slope = Δx / Δt = Vx, Av, Δt "d" very small Δ Vx = dx/dt Velocity (instantaneous) [m/s] d/dt (At^n) = An*t^(n-1) Example: x = 3t^2 - 5 = 3t^2 - 5t^0 Vx = 3.2.t^(2-1) - 0 = 6t Vx2 - Vx1 ax, Avg = --------- [m/s^2] t2 - t1 Velocity at t=4s Vx = 6.4 = 24 m/s at t=6s Vx = 6.6 = 36 m/s Example: ax, Avg = 36-24 / 6-4 = 6 m/s^2 Slope = a x Avg = tanα d -- (6t) = 6 m/s^2 dt ax = dvx/dt ← acceleration derivative Vx = dx/dt ← velocity derivative x = ∫ Vx dt Vx = ∫ ax dt ∫ tn dt = tn+1 / (n+1) + C integration constant Example: Vx = 4t^2 - 5 n = -1: ∫ dt/t = ln|t| + C Vx: 4t^2-5 x= 4t^3/3 - 5t + C to determine C, we need additional information about position at some time. (example) assume that at t=2s X=10 m 10= 4.2^3/3 - 5.2 + C C= 10- 32/3 + 10 = 3/3 - 32/3 = 28/3 Motion With constant acceleration => Vx = Vox + axt Ax-constant Vx=∫ a_{x}dt=∫ dt x=∫ v_{x}dt=∫(v_{x}+a_{x}t)dt initial position (Position x=0) (integration constant) II) X=Xo+Voxt + 1/2 axt^2 III) Calculate t from I): and Substitute to II) Vx^2=V_{0}x^{2}+2ax(x-x_{0}) END OF LECTURE Quiz given to class at the end

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