Assuming that only air resistance and gravity act on a falling object, we can find that the velocity of the object, v, must obey the differe

Question

Assuming that only air resistance and gravity act on a falling object, we can find that the velocity of the object, v, must obey the differential equation dv m mg bv dt   . Here, m is the mass of the object, g is the acceleration due to gravity, and b > 0 is a constant. Consider an object that has a mass of 100 kilograms and an initial velocity of 10 m/sec (that is, v(0) = 10). If we take g to be 9.8 m/sec2 and b to be 5 kg/sec, find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.

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Hồng Cúc 3 years 2021-09-04T12:16:04+00:00 1 Answers 4 views 0

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  1. Calantha
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    2021-09-04T12:17:30+00:00
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    Answer:

    v = 196 – 186*e^( – 0.05*t )      

    v-terminal = 196 m/s

    Explanation:

    Given:

    – The differential equation for falling object velocity v in gravity with air resistance is given by:

                                    m*dv/dt = m*g – b*v

    – The initial conditions and constants are as follows:

                     v(0) = 10 , m = 100 kg , b = 5 kg/s , g = 9.8 m/s^2  

    Find:

    – Find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.

    Solution:

    – Rewrite the differential equation in te form:

                                     dv/dt + (b/m)*v = g

    – The integration factor function P(t) = b/m. The integrating factor u(t) is:

                                     u(t) = e^∫P(t).dt

                                     u(t) = e^∫(b/m).dt

                                     u(t) = e^[(b/m).t]

    – Solve the differential equation after expressing in form:

                                     v.u(t) = ∫u(t).g.dt    

                                     v.e^[(b/m).t] = g*∫e^[(b/m).t].dt    

                                     v.e^[(b/m).t] = g*m*e^[(b/m).t] / b + C

                                      v = g*m/b + C*e^[-(b/m).t]                

    – Apply the initial conditions v(0) = 10 m/s and evaluate C:

                                      10 = 9.8*100/5 + C*e^[-(b/m).0]

                                      10 = 9.8*100/5 + C

                                      C = -186

    – The final ODE solution is:

                                      v = 196 – 186*e^( – 0.05*t )

    – The Terminal velocity vt can be expressed by a limiting value for v(t), where t ->∞.        

                                      vt = Lim t ->∞ ( v(t) )

                                      vt = Lim t ->∞ ( 196 – 186*e^( – 0.05*t ) )

                                      vt = 196 – 0 = 196 m/s

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