 ## Question:

When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity ${ v }_{ f }$. If this variation of the acceleration can be expressed as $a =(g/{ v }^{ 2 }_{ f }) ({ v }^{ 2 }_{ f } - { v }^{ 2 })$, determine the time needed for the velocity to become $v = { v }_{ f }/2$. Initially the particle falls from rest.

## Step-by-step

$\frac { dv } { dt } = a = (\frac { g } { { v }_{ f }^{ 2 } })({ v }_{ f }^{ 2 } – { v }^{ 2 }) \\ \int_{ 0 }^{ v } \frac { dv } { { v }_{ f }^{ 2 } – { v }^{ 2 } } = \frac { g } { { v }_{ f }^{ 2 } } \int_{ 0 }^{ t } dt \\ \frac { 1 } {2{ v }_{ f }} \text{ ln } (\frac { { v }_{ f } + v } { { v }_{ f } – v })|_{ 0 }^{ v } = \frac { g } { { v }_{ f }^{ 2 } } t \\ t = \frac { { v }_{ f } } { 2g } \text{ ln } ( \frac { { v }_{ f } + v } { { v }_{ f } – v } ) \\ t = \frac { { v }_{ f } } { 2g } \text{ ln } (\frac { { { v }_{ f } } + { v }_{ f }/2 } { { v }_{ f } – { v }_{ f }/2 }) \\ t = 0.549(\frac{ { v }_{ f } } { g })$