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?

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    $\begingroup$ "To obtain s, we multiply 10.5=50 ." - where does it come from? $\endgroup$ – Dmitry Aug 1 at 13:54
  • $\begingroup$ @Dmitry, t=5, at t = 5, v = 10, so s = 10 . 5 = 50 $\endgroup$ – wawar05 Aug 1 at 13:56
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    $\begingroup$ $s = v \times t$ is a formula for constant speed. When it changes, you need to take an integral, like you did. $\endgroup$ – Dmitry Aug 1 at 13:58
  • $\begingroup$ @Dmitry so, my solution for the problem is correct $\endgroup$ – wawar05 Aug 1 at 14:00
  • $\begingroup$ 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. $\endgroup$ – 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.

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