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Can you do part A and B only I don’t understand in A how they got v’-vT I’m getting the vice versa and I why is there a negative in the intergration

Can you do part A and B only I don’t understand in A how they got v’-vT I’m getting the vice versa and I why is there a negative in the intergration

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Can you do part A and B only I don’t understand in A how they got v’-vT I’m getting the vice versa and I why is there a negative in the intergration
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 FD = -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? Problem 2-2D Collisions marr me Before the collisions, particle B is at rest, while particle A moves along the x-axis
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 FD = -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? Problem 2-2D Collisions marr me Before the collisions, particle B is at rest, while particle A moves along the x-axis
01 Problem 1- (a) m Newton's I law: -bv ma V₁ = b dv dt = mg- It is convenient to introduce the terminal velocity 1 ка x 10 m/s² 10 kg/s and the drag time. 5 = 1/² Thus : dv dt V V₂ = V4 TD dv' V'- VT ln V-VT Vo-VT = 1D gravity and drag 1kg 10 - ka/s V - JO t هانا = 10 s 2/8 dt' ZD = I 100 m/s 91% + V-V7 V-VT VT v(t)= V₁ +(V-V₁)e V^ Vo O -t/TD -t/TD =. TD (b) Find y(t) by integrating dy dt -t/tp V(t) = +Vo-V+) e -t/D y(t) = + (-₂) e + G y (0)=0 → C=(V-VT) T y(t) = vt + (₁-v₂) (e ths-1) For t= 10s = p
Transcribed Image Text:01 Problem 1- (a) m Newton's I law: -bv ma V₁ = b dv dt = mg- It is convenient to introduce the terminal velocity 1 ка x 10 m/s² 10 kg/s and the drag time. 5 = 1/² Thus : dv dt V V₂ = V4 TD dv' V'- VT ln V-VT Vo-VT = 1D gravity and drag 1kg 10 - ka/s V - JO t هانا = 10 s 2/8 dt' ZD = I 100 m/s 91% + V-V7 V-VT VT v(t)= V₁ +(V-V₁)e V^ Vo O -t/TD -t/TD =. TD (b) Find y(t) by integrating dy dt -t/tp V(t) = +Vo-V+) e -t/D y(t) = + (-₂) e + G y (0)=0 → C=(V-VT) T y(t) = vt + (₁-v₂) (e ths-1) For t= 10s = p
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where is the explanation for part b??

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