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Could you explain how they are integrating the equations for this question please

Could you explain how they are integrating the equations for this question 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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Could you explain how they are integrating the equations for this question please
Problem 1-1D Gravity and Drag A particle with mass m = 1 kg falls along the vertical y direction and is subject to a drag force Fo= -bv, with b=0.1 kg/s. Assume the y-axis directed downward, so that the force of gravity is positive. The velocity of the particle at time t=0 is also along the y-direction, with positive (downward) value v(0) = Vo = 200 m/s, and the particle initial height is y(0) = 0. (a) Integrate Newton's second law and find the particle velocity v(t) as a function of time. Sketch the graph v(t). (b) Find the particle position y(t) as a function of time. Estimate the particle's position at time t= 10 s (Hint: e¹¹ 0.37; in this problem, you can use g = 10 m/s2). (c) Consider the special case vo= 100 m/s. What happens in this case? (a) Newton's I law: mdv = mg-bu, it is convenient to introduce terminal velocity and the drag time is: | time is: to = m = 1.1 = 10₁ ĭ dv² = -√ dh V₁ = m² = 1x 10 = 100 m/s 0.1 Thus: du = VT - 2/10 So > v(t) = V₁ + (vo-v₁) et/% O ln t == Vo -VT I V-VT -t/to (b) Find yit) by integrating dy=v(t) = V₁ + (v₁ - v₁) ²²₂ y(t) = v₁t+ (v₁-v₂)/² + glo)=0⇒c= (vo-v7) y(t) = vyt+ (v₂-v₂) (etto -1)
Transcribed Image Text:Problem 1-1D Gravity and Drag A particle with mass m = 1 kg falls along the vertical y direction and is subject to a drag force Fo= -bv, with b=0.1 kg/s. Assume the y-axis directed downward, so that the force of gravity is positive. The velocity of the particle at time t=0 is also along the y-direction, with positive (downward) value v(0) = Vo = 200 m/s, and the particle initial height is y(0) = 0. (a) Integrate Newton's second law and find the particle velocity v(t) as a function of time. Sketch the graph v(t). (b) Find the particle position y(t) as a function of time. Estimate the particle's position at time t= 10 s (Hint: e¹¹ 0.37; in this problem, you can use g = 10 m/s2). (c) Consider the special case vo= 100 m/s. What happens in this case? (a) Newton's I law: mdv = mg-bu, it is convenient to introduce terminal velocity and the drag time is: | time is: to = m = 1.1 = 10₁ ĭ dv² = -√ dh V₁ = m² = 1x 10 = 100 m/s 0.1 Thus: du = VT - 2/10 So > v(t) = V₁ + (vo-v₁) et/% O ln t == Vo -VT I V-VT -t/to (b) Find yit) by integrating dy=v(t) = V₁ + (v₁ - v₁) ²²₂ y(t) = v₁t+ (v₁-v₂)/² + glo)=0⇒c= (vo-v7) y(t) = vyt+ (v₂-v₂) (etto -1)
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