Integrating velocity at a time into distance at a time

**Karl Harder** · September 17th 2009, 03:36 AM

I am trying to integrate this with respect to time - t
Note: t is equal to 600-t

$v(t) = -35-100 \frac {dv}{dt}$

I tried to integrate it and got this

$d(t) = -35(600-t)-100v(t)$

somehow i it should be

$d(t) = 100v(t)-35(600-t)$

**The Second Solution** · September 17th 2009, 03:51 AM

Originally Posted by Karl Harder

I am trying to integrate this with respect to time - t
Note: t is equal to 600-t

$v(t) = -35-100 \frac {dv}{dt}$

I tried to integrate it and got this

$d(t) = -35(600-t)-100v(t)$

somehow i it should be

$d(t) = 100v(t)-35(600-t)$

I have no idea what

Note: t is equal to 600-t

means. At any rate,

$v = -35 - 100 \frac{dv}{dt} \Rightarrow \frac{v + 35}{100} = - \frac{dv}{dt} \Rightarrow \frac{dt}{dv} = - \frac{100}{v + 35}$ .

Therefore $t = - \int \frac{100}{v + 35} \, dv$ etc.

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