# Thread: Solution to a distance travelled problem

1. ## Solution to a distance travelled problem

I have the following question in an assignment,

A particle moves along a straight line so that at time t seconds its velocity, is given by

$v=\frac{t+1}{t^2+1}$

Find the distance travelled in the first second.

Solving the equation, I got a general solution to the differential equation as:

$s(t)=\frac{1}{2} ln(t^2+1) + tan^{-1} t + C$

Is it correct to assume that the initial condition would be s=0 at t=0?

I went on to solve the problem as s = 1.132m.

Can anyone check this for me and show me where I went wrong if there is a problem?

2. ## Re: Solution to a distance travelled problem

Originally Posted by tammyl
I have the following question in an assignment,

A particle moves along a straight line so that at time t seconds its velocity, is given by

$v=\frac{t+1}{t^2+1}$

Find the distance travelled in the first second.

Solving the equation, I got a general solution to the differential equation as:

$s(t)=\frac{1}{2} ln(t^2+1) + tan^{-1} t + C$

Is it correct to assume that the initial condition would be s=0 at t=0?

I went on to solve the problem as s = 1.132m.

Can anyone check this for me and show me where I went wrong if there is a problem?
That's the answer. As to s = 0 where t = 0 you are looking for how far it went. So if you don't choose s = 0 then you have to subtract the value to get rid of the extra difference. (If s = 1, t = 1 gives the distance as 2.132 m, and to get the actual distance traveled you need to do $\Delta s$ - 1 = 1.132.)

-Dan

3. ## Re: Solution to a distance travelled problem

You can certainly assume that the "first second" is from t=0 to t= 1. You do NOT have to assume that s(0)= 0 since s(0)= C will give the same answer for any C.