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I don’t understand how they’re getting the formula of Vx(t) at the start by integrating could you show me how to resolve this in steps please

I don’t understand how they’re getting the formula of Vx(t) at the start by integrating could you show me how to resolve this in steps please

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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I don’t understand how they’re getting the formula of Vx(t) at the start by integrating could you show me how to resolve this in steps please
Problem 1-2D Kinematics and drag. * An object having mass m = 1 kg is launched with initial velocity at t=0, v(0) = Vxoi + Vyoj, with Vxo = Vyo = 200 m/s, and initial position x(0) = y(0) = 0. The forces acting of the object are gravity, Fg = -mg j, where you can approximate g = 10 m/s², and drag, Fo= -bv, where b = 0.1 kg/s. (a) Find the time, tmax, it takes for the object to reach maximum height; (b) Find the velocity along the x-axis at maximum height, vx(tmax); (c) Find the value of the maximum height, ymax=y(tmax). (Hint: use In3 = 1.1). (a) The equation of motion (Newton's Il law) V m ,dvx = -bvx ⇒v₂(t)=v₂.e ² at M (mduy = -mg-buy -witten Exam 3 September 2021 Solving for tmox: tmax = To bn (1 + 1/²) tmax = 10n3=11s (b) At t=tmox, v₂ (tmax) = √₂e tmax/20 v₂ (tmax)= V₂o = 67 m/s 1+ yo VT →200 1+200 J00 The solution of y-component is: - VT Vy(t) = (V₁ + Vyo)et/to_ where To = 1/ = 1 = 10 m VT = mg = 1x/0= 100 m/s (c) We can obtain an expression for yit) by integrating d y =Vylt) = (V₁ + Vyo) e t/to_ we obtain: y(t)= -(v +vy)e t/to -v₁t+c where c is an integration constant At t=0, ylo) = C-²₂ (V₁ + Vy₂) = 0 and so c= (v₁+₁)%o and
Transcribed Image Text:Problem 1-2D Kinematics and drag. * An object having mass m = 1 kg is launched with initial velocity at t=0, v(0) = Vxoi + Vyoj, with Vxo = Vyo = 200 m/s, and initial position x(0) = y(0) = 0. The forces acting of the object are gravity, Fg = -mg j, where you can approximate g = 10 m/s², and drag, Fo= -bv, where b = 0.1 kg/s. (a) Find the time, tmax, it takes for the object to reach maximum height; (b) Find the velocity along the x-axis at maximum height, vx(tmax); (c) Find the value of the maximum height, ymax=y(tmax). (Hint: use In3 = 1.1). (a) The equation of motion (Newton's Il law) V m ,dvx = -bvx ⇒v₂(t)=v₂.e ² at M (mduy = -mg-buy -witten Exam 3 September 2021 Solving for tmox: tmax = To bn (1 + 1/²) tmax = 10n3=11s (b) At t=tmox, v₂ (tmax) = √₂e tmax/20 v₂ (tmax)= V₂o = 67 m/s 1+ yo VT →200 1+200 J00 The solution of y-component is: - VT Vy(t) = (V₁ + Vyo)et/to_ where To = 1/ = 1 = 10 m VT = mg = 1x/0= 100 m/s (c) We can obtain an expression for yit) by integrating d y =Vylt) = (V₁ + Vyo) e t/to_ we obtain: y(t)= -(v +vy)e t/to -v₁t+c where c is an integration constant At t=0, ylo) = C-²₂ (V₁ + Vy₂) = 0 and so c= (v₁+₁)%o and
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