# Distance as the integral of velocity problem

I am making my own example of the application of integral in physics for my task as follows:

the velocity of an object at the time $$t$$ is given by $$v(t)=2t$$, how far has the object traveled at $$t=5$$?

my solution is that I integrate the given velocity function $$2t$$ resulting in a distance function $$s(t)=t^2$$ from $$t=0$$ to $$t=5$$ which is equal to 25.

Is that correct?

If it is, what about we substitute $$t$$ with 5 in the function $$v(t)=2t$$ which makes $$v=10$$. To obtain $$s$$, we multiply $$10.5=50$$ .

perhaps, there is something wrong either with my example or my second thought, but what does make it different?

• "To obtain s, we multiply 10.5=50 ." - where does it come from? – Dmitry Aug 1 at 13:54
• @Dmitry, t=5, at t = 5, v = 10, so s = 10 . 5 = 50 – wawar05 Aug 1 at 13:56
• $s = v \times t$ is a formula for constant speed. When it changes, you need to take an integral, like you did. – Dmitry Aug 1 at 13:58
• @Dmitry so, my solution for the problem is correct – wawar05 Aug 1 at 14:00
• To dramatize @Dmitry 's point, using $s = vt$ for this situation is like using the simple formula for the area of a rectangle ($A = b \cdot h$) and applying it to a triangle. You just won't get the right answer. – JonathanZ supports MonicaC Aug 1 at 14:02

Your first approach is correct. Indeed, by definition, one has: $$\begin{eqnarray} v = \frac{ds}{dt} \Rightarrow s(t)-s(t_{0}) = \int_{t_{0}}^{t}v(t)dt \tag{1}\label{1} \end{eqnarray}$$ The second approach, where you take $$t=5$$, so that $$v(5) = 10$$ and then multiply it by $$5$$ is not correct. I believe your reasoning to do that is to consider $$v=\frac{\Delta s}{\Delta t}$$, so that $$\Delta s = v \Delta t$$. But note that this formula only holds when $$v$$ is assumed to be constant. The correct definition of $$v$$ is given by (\ref{1}) and if $$v$$ is constant, we have: $$\int_{t_{0}}^{t}v(t)dt = v\int_{t_{0}}^{t} = v(t-t_{0})= v\Delta t$$ from where it follows that $$s(t)-s(t_{0}) = \Delta s = v\Delta t$$. But if $$v$$ is a function of $$t$$, as it is in your particular case, this formula does not hold anymore unless you'd like to compute the average velocity of the particle.