Calculating distance when velocity is given

Question

How to find the distance in $0 < t < 2$ when the velocity is given:
$$v(t) = frac{t^3}{10} - frac{t^2}{20} + 1$$
I have tried the following and got the answer as $2.267$
begin{align*}
d = int_{0}^{2}v(t)mathrm{d}t & Longleftrightarrow d = int_{0}^{2} left(frac{t^{3}}{10} - frac{t^{2}}{20}+1right)mathrm{d}t\
& Longleftrightarrow d = left(frac{t^{4}}{40} - frac{t^{3}}{60}+tright)Big|_{0}^{2}\
& Longleftrightarrow d = 2.267 m
end{align*}
Is this approach correct? Kindly help.

user1337 · Answer

There is a distinction between the difference of the particle's position between the instants $t_{2} > t_{1}$ and the distance traveled by the particle corresponding to the time interval $Delta t = t_{2} -t_{1}$. The first is given by the integral of $v(t)$ and the second is given by the integral of $|v(t)|$. Depending in which case you are interested, you can calculate the corresponding integral in order to obtain the answer.

Answered by user1337 on July 27, 2020

Calculating distance when velocity is given

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