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thejohan

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Is what I have done correct ?

I want to find v(t) from Sigma F = m*a. I have gravity force mg pointing downward with positive direction and resistive force R = -b*v^2 pointing upwards with negative direction are acting on a particle. So

( a = dv / dt) , m ,b and g are constants

I start with : Sigma F = m *a

mg - b*v^2 = m*a

g - b/m * v^2 = a

dv/dt + b/m v^2 = g

(dv/dt) / (g - (b/m) * v ^2) = 1

(1 / g - (b/m) * v ^2) * (dv/dt)= 1

integral (1 / g - (b/m) * v ^2) * (dv/dt) * dt = integral 1 dt

I use substitution u = g - (b/m) * v^2

du/ dt = -(b/m) * v

du =(2b/m)* v * dt

(2b/m)* v * du = dt

(2b/m)*v integral 1 / u du = integral 1 dt

(2b/m ) * v * ln |u| = t + c

(2b/m ) * v * ln |g - (b/m) * v^2 | = t +c

Is what I have done correct ?

I want to find v(t) from Sigma F = m*a. I have gravity force mg pointing downward with positive direction and resistive force R = -b*v^2 pointing upwards with negative direction are acting on a particle. So

( a = dv / dt) , m ,b and g are constants

I start with : Sigma F = m *a

mg - b*v^2 = m*a

g - b/m * v^2 = a

dv/dt + b/m v^2 = g

(dv/dt) / (g - (b/m) * v ^2) = 1

(1 / g - (b/m) * v ^2) * (dv/dt)= 1

integral (1 / g - (b/m) * v ^2) * (dv/dt) * dt = integral 1 dt

I use substitution u = g - (b/m) * v^2

du/ dt = -(b/m) * v

du =(2b/m)* v * dt

(2b/m)* v * du = dt

(2b/m)*v integral 1 / u du = integral 1 dt

(2b/m ) * v * ln |u| = t + c

(2b/m ) * v * ln |g - (b/m) * v^2 | = t +c

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