So we have velocity as a function of time given $\overrightarrow{v}(t)=(1-2bt)k\overrightarrow{r_0}$ where $b,k$ are positive constants and $\overrightarrow{r_0}$ is a unit vector. It is found that if $t=\frac1b$ then $\overrightarrow{r_0}=0$. It is asked to find the distance traveled for $\Delta{t}=\frac1b$. At first I thought to apply the general formula for the length of a curve which in this case would be $$\int_a^b{\sqrt{1+{v}(t)^2}dt}=\int_{0}^{1/b}{\sqrt{1+{v}(t)^2}dt}=\frac{(k+1)^{\frac32}-(1-k)^{\frac32}}{3kb}$$ However in the textbook this is done by calculating $$\int_0^tvdt=\int_0^{\frac1{2b}}(1-2bt)kdt+\int_{\frac1{2b}}^{\frac1b}(1-2bt)kdt=\frac{k}{2b}$$ However by hand calculation I've obtained $$\int_0^tvdt=\int_0^{\frac1{2b}}(1-2bt)kdt+\int_{\frac1{2b}}^{\frac1b}(1-2bt)kdt=0$$ which I would guess gives displacement but not the path traveled. So these are my questions:

  1. Is the approach using the general formula for the length of a curve correct for this kind of problem, please explain?
  2. Is the result obtained by the approach in the textbook correct, and why is this approach correct since I would say it gives the displacement and not the distance traveled?
  • $\begingroup$ Hi Ahmed and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Jul 7 '16 at 9:44
  • $\begingroup$ In a 1D problem like this the total distance moved is obtained by integrating the modulus of the velocity i.e. $\int\,|v(t)|\,dt$. The book does this by splitting the integral into two parts, one where $\vec{v}$ is positive and the other where $\vec{v}$ is negative. Your second equation should have a $-$ sign between the two integral or the second integral should be $\int(2bt-1)kdt$. $\endgroup$ – John Rennie Jul 7 '16 at 9:48

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