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The position vector of a particle is given as $\vec r = \frac43 t^{3/2}\hat i - \frac{1}{2} t^2\hat j + 2 \hat k$, $t$ is in seconds. Find the distance travelled between $t = 0$ and $t = 5$ seconds.

I tried it solving by putting values of $t$ as $0$ and $5$ seconds in the position vector, but I couldn't get the answer. (In reality, when I put $t=5$, the equation got really messy.) Please help me. Any help will be appreciated.

The given answer is $22.5$ m.

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You need to show some work.

I think your position equation should be $$\vec r(t) =\left(\frac{4}{3}t^{\frac{3}{2}}\right)\hat\imath +\left(-\frac{t^{2}}{2}\right)\hat\jmath +\left(2\ \underline{t}\ \right)\hat k$$ in order to get the value listed in your spoiler.

It's generally a good idea to start with the definitions.

  • What is distance? (Don't confuse with "displacement" or with "magnitude of displacement".)
  • How do you compute distance traveled when is the motion isn't steady motion?
  • What parts do you need to compute the distance?
  • How can you compute the needed parts from what you are given?
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  • $\begingroup$ I will try this. Thanks for the Effort! : ) $\endgroup$
    – Mritunjay Kumar
    Jun 19, 2021 at 10:19

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