nagle_differential_equations_ISM_Part20

nagle_differential_equations_ISM_Part20 - Exercises 3.4...

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Unformatted text preview: Exercises 3.4 Therefore, solving x 1 ( t ) = 1 . 962 t- . 392 ( 1- e- 5 t ) = 30 , we obtain t ≈ 15 . 5 sec for the time when the object hit the water. The velocity of the object at this moment was v 1 (15 . 5) = 1 . 962 ( 1- e- 5(15 . 5) ) ≈ 1 . 962 . We now go to the motion of the object in the water. For convenience, we reset the time. Denoting by x 2 ( t ) the distance passed by the object from the water surface and by v 2 ( t ) – its velocity at (reset) time t , we get we obtain initial conditions v 2 (0) = 1 . 962 , x 2 (0) = 0 . For this motion, in addition to the gravity force F g = mg and the resistance force F r =- 100 v , the buoyancy force F b =- (1 / 2) mg is presented. Hence, the Newton’s second law yields m dv 2 dt = mg- 100 v- 1 2 mg = 1 2 mg- 100 v ⇒ dv 2 dt = g 2- 100 m v 2 = 4 . 905- 50 v 2 . Solving the first equation and using the initial condition yields v 2 ( t ) = 0 . 098 + Ce- 50 t , v 2 (0) = 0 . 098 + C = 1 . 962 ⇒ C = 1 . 864 ⇒ v 2 ( t ) = 0 . 098 + 1 . 864 e- 50 t ⇒ x 2 ( t ) = Z t v 2 ( s ) ds = 0 . 098 t- . 037 e- 50 t + 0 . 037 . Combining the obtained formulas for the motion of the object in the air and in the water and taking into account the time shift made, we obtain the following formula for the distance from the object to the platform x ( t ) = 1 . 962 t- . 392 (1- e- 5 t ) , t ≤ 15 . 5 . 0981( t- 15 . 5)- . 037 e- 50( t- 15 . 5) + 30 . 037 , t > 15 . 5 . 1 min after the object was released, it traveled in the water for 60- 15 . 5 = 44 . 5 sec....
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nagle_differential_equations_ISM_Part20 - Exercises 3.4...

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