You first need to know that velocity is derivation of distance traveled over time, that is

`v(t)=(ds(t))/(dt)`

**a)**

So if we have velocity `v` then distance traveled is integral of velocity

`s(t)=int_0^t v(x)dx`

`s(t)=int_0^tdx/(1+x)=(ln|1+x|)|_0^t=ln(1+t)-ln1=ln(1+t)`**<--Solution**

We have `ln(1+t)` instead of `ln|1+t|` because `t` is always non-negative so `ln|1+t|=ln(1+t)`.

**b)**

It is easy to see that `s(t)` is monotonically increasing function because `ln(1+t)` is increasing...

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You first need to know that velocity is derivation of distance traveled over time, that is

`v(t)=(ds(t))/(dt)`

**a)**

So if we have velocity `v` then distance traveled is integral of velocity

`s(t)=int_0^t v(x)dx`

`s(t)=int_0^tdx/(1+x)=(ln|1+x|)|_0^t=ln(1+t)-ln1=ln(1+t)`**<--Solution**

We have `ln(1+t)` instead of `ln|1+t|` because `t` is always non-negative so `ln|1+t|=ln(1+t)`.

**b)**

It is easy to see that `s(t)` is monotonically increasing function because `ln(1+t)` is increasing function. In fact all logarithms with base greater than 1 are increasing functions. Formally you can prove it like this:

`s'(t)=(ln(1+t))'=1/(1+t)`

and since `t>0` we have `s'(t)>0` which means that `s` is monotonically increasing function.

**c)**

Here you should know that acceleration is derivation of velocity over time (or second derivation of distance traveled over time) that is

`a(t)=(dv(t))/(dt)`

`a(t)=(1/(1+t))'=-1/(1+t)^2`

So at time `t=0` acceleration is

`a(0)=-1/(1+0)^2=-1` **<--Solution**