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The displacement $s$ metres of an object is given by $s(t)=e^{\cos(t)},$ where $0 \leq t \leq \pi$ and the time is $t$ seconds.

Determine the displacement of the function at $\frac{\pi}{2}$ seconds.

I computed $s(\frac{\pi}{2})-s(0)$ and come up with $1-e$, but my tutor is insisting that the displacement for this question is relative to the origin, i.e., the solution is $1$.

Am I wrong in assuming that I need to find the displacement at time equals 0 and then the displacement relative to that point. Is this a different part of math that I just haven't learnt yet?

ryang
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BJH
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  • Welcome to stackexchange!

    This sounds like an opinion driven question that may not belong here. You could argue like you, like the tutor, against the global minimal displacement...

     – Michael T Mar 03 '25 at 12:29
  • not a perfect fit to the site but possibly of interest on what to compare to in a dynamics question... – Michael T Mar 03 '25 at 12:30
  • @MichaelT Thanks Michael. I guess the fault lies in the ambiguity of the question then? – BJH Mar 03 '25 at 12:39
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    Displacement is defined as the difference between the final and initial positions of an object. – N. F. Taussig Mar 03 '25 at 12:44
  • @BJH, well, if you believe you can win that argument against your tutor, you can try... ;-) – Michael T Mar 03 '25 at 12:46
  • @N.F.Taussig I have assumed initial position to be $e$, of which is already a displacement of $e$ from the origin... but this is being refuted by saying that its original position is the origin. – BJH Mar 03 '25 at 12:52
  • @MichaelT I tried and it died hard. His case in point was lets simplify it "Given a function $s(t)=t+10$ what is its displacement at $t=0$ ... $10$ to which he agreed but, then what about when $t=1$... I said $1$ but apparently its $11$. So maybe he is looking at it from the perspective of absolute in space? Unfortunately I don't think I have the knowledge to try and argue this case and he can mark me as wrong... lol. Thank you for you assistance :) – BJH Mar 03 '25 at 12:56
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    @BJH I agree with your answer based on the definition of displacement. – N. F. Taussig Mar 03 '25 at 13:10
  • I agree with OP. – Adam Rubinson Mar 03 '25 at 14:30
  • Just to clarify a little. Does the statement say that $s(t)$ is the displacement at time $t$, or does it say that $s(t)$ is the position at time $t$? If it's the former, then your tutor is correct, sort of by definition, although that makes the problem really strange, and I would never write a problem like this; if this is the case, then I think the problem is with the problem, not you or the tutor. On the other hand, if $s(t)$ is the position, then absolutely the displacement would be $s(t_f) - s(t_i)$. – march Mar 04 '25 at 05:03

3 Answers3

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In general cases, I would agree with Michael T’s answer. However, your problem states that the function $s(t)$ models the displacement of an object with time. This means that if you pick any point on the graph, the $y$-value of that point already is the displacement of the object at that time. There’s no need to do additional operations by comparing it to $s(0)$. It’s like having a function $f(t)$ of an item’s price over time, and if you wanted to find the item’s price at $t=4$, you would simply find the value of $f(4)$ instead of doing $f(4)-f(0)$. This second expression would only tell you the change in price over the time interval, but not the actual price. Similarly, your method of doing $s\left(\frac{\pi}{2}\right)-s(0)$ only tells you the change in displacement over the time interval, but not the actual displacement at $t=\frac{\pi}{2}$.

You can take it a step further by inferring that if the $y$-values of the function tell you the displacement, then the problem assumes displacement to be relative to the $x$-axis, since the displacement of any point from the $x$-axis is represented by the $y$-coordinate of that point.

VV_721
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1

The displacement $s$ metres of an object is given by $\boldsymbol {s(t)=e^{\cos(t)}}$

Determine the displacement of the function at $\boldsymbol{\frac{\pi}{2}}$ seconds.

I computed $s(\frac{\pi}{2})-s(0)$ and come up with $1-e$, but my tutor is insisting that the displacement for this question is relative to the origin, i.e., the solution is $1$.

The given instruction is just straightforwardly asking for the value of $$s\left(\frac{\pi}2\right).$$

On the other hand, your computation $$s\left(\frac{\pi}{2}\right)-s(0)$$ corresponds to the instruction, "Determine the object's displacement over the first $\boldsymbol{\frac{\pi}2}$ seconds."

For clarity, I would call $s$ the object's position (relative to the origin), or its displacement from the origin, instead of merely its displacement (i.e., change in position).

ryang
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The way the exercise is put ('displacement $s$ [in] metres of an object') is clear and asks to calculate $s(\pi/2)$.

However, you could argue there is some room for a discussion of the most reasonable frame of reference to compare to

  • absolute in space: $s=0$
  • absolute in time: $t=0$, ie $s=e$
  • minimum of $s$: $1/e$
  • average of $s$ in space: $s=1.26606...$ = $\frac{1}\pi \int_0^{\pi} e^{\cos(t)}$
  • other?

Of course, that is bit like 'shooting at sparrows w cannons'.

Michael T
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