Distance travelled by a falling body

• Bl4nk
In summary, the first second of a 4 second moving object covers 2 meters, the second covers 3 meters, and so on.
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.

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

Bl4nk and topsquark
haruspex said:
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

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

Bl4nk and topsquark
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.

Bl4nk
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.

What is the formula for calculating the distance travelled by a falling body?

The formula for calculating the distance travelled by a falling body is d = 1/2 * g * t^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds.

Does the mass of the falling body affect the distance travelled?

Yes, the mass of the falling body does affect the distance travelled. The heavier the object, the greater the force of gravity acting on it, and therefore the greater the distance it will travel in a given amount of time.

What is the difference between distance and displacement?

Distance refers to the total length of the path travelled by an object, while displacement refers to the straight-line distance between the starting point and ending point of the object's motion.

How does air resistance affect the distance travelled by a falling body?

Air resistance, also known as drag, can decrease the distance travelled by a falling body. As the object falls, it experiences an upward force from the air pushing against it, which can slow it down and reduce the distance it travels.

What other factors can affect the distance travelled by a falling body?

In addition to mass and air resistance, the height from which the object is dropped, the shape and size of the object, and the presence of external forces (such as wind or air currents) can also affect the distance travelled by a falling body.

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