Mathematics Asked by Aruha on July 27, 2020

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.

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

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