Distance travelled by a falling body

[Bl4nk]

Homework Statement

"For a freely falling body from rest, not counting air resistance or any frictional force, between the third and fourth second of time, it travels a distance of?"
What does the question mean?
a. Distance traveled in the 4th second.
or
b. The difference between the distances travelled in the 4th second and the 3rd second.
And why?

Relevant Equations

h = ut + 1/2gt^2

If (a) is correct, the answer would be 34.3m.
If (b) is correct, the answer would be 9.8m.
I want to know exactly which one is meant by the question and the reason behind that. Personally i think its a.

 

[haruspex]

Yes, it's a. I don’t see how you could read it as b. Though I do prefer your rewording.

 

[Bl4nk]

Yes, it's a. I don’t see how you could read it as b. Though I do prefer your rewording.

Apparently I couldn't make one of my friends understand that its the question asks for A and not B

 

[DrClaude]

Mathematically it asks for the distance $d = \left| h(t=4 \mathrm{s}) - h(t=3 \mathrm{s}) \right|$

 

[jbriggs444]

One could argue that the correct answer is zero. The wording of the question is atrocious.

The "third second" starts at $t=2$ and ends just prior to $t=3$. The "fourth second" starts at $t=3$ and ends just prior to $t=4$. Between the two intervals, zero time elapses and zero distance is traversed.

[The "first second" starts at $t=0$ of course].

Possibly the intent is that "first second" denotes the instant at which the object is released at $t=0$.
Possibly the intent is that "first second" denotes the instant one second later at $t=1$.

Three plausible readings is an indication of poor writing.

 

[kuruman]

I agree with @DrClaude's interpretation. Yes, the statement could have been clearer, but it is clear to me that there is an assumed imaginary clock that starts when motion starts, as is usually the case. Then one counts ticks of the second hand, tick 1, tick 2, etc. and figures out distances $d_1$, $d_2$, etc. covered from one tick to the next. Thus, $$\begin{align} & d_1=\frac{1}{2}a(1~\mathrm{s})^2 - \frac{1}{2}a(0~\mathrm{s})^2\nonumber \\ & d_2=\frac{1}{2}a(2~\mathrm{s})^2-\frac{1}{2}a(1~\mathrm{s})^2=3d_1 \nonumber \\ & \dots \nonumber \\ & d_n=\frac{1}{2}a \left[n^2-(n-1)^2\right]\mathrm{s}^2= (2n-1)d_1~. \nonumber \end{align}$$

 

[Bl4nk]

Thanks everyone for explaining this so elaborately. For real tysm ^w^. Helped a lot.