The particle is falling down, and x is measured in down direction. x ″ = g − k v 4 ⟹ v d v d x = g − k v 4 Solving this I get 1 4 g k log ⁡ ( k v 2 − g k v 2 + g ) = x + cIt is falling from rest, i.e x=0,v=0, using this initial condition I get log ⁡ ( − 1 ) on the LHS. Is something wrong with the solution. I check that integral is correct.

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The particle is falling down, and x is measured in down direction.      x    ″    =  g  −  k      v    4        ⟹      v            d      v              d      x        =  g  −  k      v    4  Solving this I get      1          4              g        k              log  ⁡            (                          k                    v        2            −              g                            k                    v        2            +              g                        )        =  x  +  cIt is falling from rest, i.e x=0,v=0, using this initial condition I get   log  ⁡  (  −  1  ) on the LHS. Is something wrong with the solution. I check that integral is correct.

cegielnikmzjkf · Accepted Answer

Hint: Your answer is wrong From   v            d      v              d      x        =  g  −  k      v    4   we have  ∫                    v                 d        v                    g        −        k                  v          4                      =  d  xlet substitution       v    2    =            g      k        w then  x  =            1              2                  k          g                      ∫                    d        w                    1        −                  w          2                      +  C  =            1              4                  k          g                      ln  ⁡                    1        +        w                    1        −        w              +  Cnow let x=0, v=0 then C=0

The particle is falling down, and x is measured in down direction. x ″ </

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