A particle moves along $x$-axis such that each position is given by $x(t)=2t^3-15t^2+36t+5$, $x$ is expressed in metres. Find the total distance within the time interval $t=0$ second to $t=4$ seconds.

My method: I integrated the equation $x(t)=2t^3-15t^2+36t+5$ to get $116$ metres as the answer. However, according to my sir, the answer is $34$ m. According to him the answer which I got is the displacement of the body and not distance since it depicts the area under the curve formed when a velocity-time graph is drawn.

Now my question is, if $116$ metres is the displacement and $34$ metres is the distance, how is this even possible because displacement can never be more than the distance covered by the body. Please help me. Please clear the doubt and correct my method if wrong.


closed as off-topic by sammy gerbil, Jon Custer, ZeroTheHero, Yashas, John Rennie Jun 18 '17 at 14:39

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  • 3
    $\begingroup$ integration of $x(t)$ wrt time has no physical significance. $\endgroup$ – xasthor Jun 17 '17 at 17:58

What you calculate is neither displacement or distance. The displacement is the integration of velocity. Here, $x=2t^3-15t^2+36t+5$ is the expression of displacement already. The displacement is $$x(4)-x(0)=32 $$

  • $\begingroup$ So 116m is not at all any answer, right? $\endgroup$ – Lokesh Sangewar Jun 17 '17 at 17:06
  • $\begingroup$ Yes, it has no physical meaning. $\endgroup$ – Eric Yang Jun 17 '17 at 17:07
  • $\begingroup$ And what exactly is the correct answer? $\endgroup$ – Lokesh Sangewar Jun 17 '17 at 17:08
  • 1
    $\begingroup$ The displacement is 32, the total distance is 34 $\endgroup$ – Eric Yang Jun 17 '17 at 17:09

When you plot the given parametric equation, you see the following graph. The equation already tells you how to find the distance t any given time, so first of all, use that information to your advantage. $$x(4)=37$$ $$x(0)=5$$ $$x(4)-x(0)=32$$ observe from the graph, that as t approaches the 2 second mark, the object begins to travel in the $-x$ direction. You need to take that into account.


You can find the t-values for the maxima ($t_1$) and minima ($t_2$) by differentiating $x(t)$ with respect to $t$, setting the answer to zero and finding the roots.

Find and combine:

  • $ \left( x(t_2)-x(t_1) \right)$
  • $x(4)-x(0)$
  • $\begingroup$ Yh I understood that, also the velocity becomes zero at 2 and 3 second $\endgroup$ – Lokesh Sangewar Jun 17 '17 at 17:07
  • $\begingroup$ Do you now understand the problem, and are you able to demonstrate this to your teacher? $\endgroup$ – DWD Jun 17 '17 at 17:12

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