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The question:

An object is dropped from a cliff. How far does the object fall in the 3rd second?"

I calculated that a ball dropped from rest from a cliff will fall $45\text{ m}$ in $3 \text{ s}$, assuming $g$ is $10\text{ m/s}^2$.

$$s = (0 \times 3) + \frac{1}{2}\cdot 10\cdot (3\times 3) = 45\text{ m}$$

But my teacher is telling me $25\text{ m}$!

EDITS: His reasoning was that from $t=0$ to $t=1$, $s=10\text{ m}$, and from $t=1$ to $t =2$, $s=20$...

The mark scheme also says $25\text{ m}$

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    $\begingroup$ Please use MathJax to format your equation. In this case, e.g.$$y(t) = \frac{1}{2} a t^2$$With $a = 10 m/s^2$, $y(3s) = 45m$ indeed. Just because someone in authority claims something, does not make that right. After showing your work, and they claim a different answer, you should ask them for the proof. If they refuse, they are not a "teacher" but an indoctrination specialist. $\endgroup$ – Nominal Animal Apr 2 '17 at 10:04
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    $\begingroup$ @projectilemotion "an object is dropped from a cliff. How far does the object fall in the 3rd second?" $\endgroup$ – ChubbyChoc Apr 2 '17 at 10:16
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    $\begingroup$ @ChubbyChoc: The teacher and the mark scheme are wrong and you right. Ignoring air resistance, with the object is stationary before dropping (initial velocity is zero), the only force acting on it is Earth gravity, which is between $9.78 m/s^2$ and $9.83 m/s^2$ depending on the location (and is basically constant in the region an initially stationary object can reach in three seconds without other forces acting upon it). The displacement due to constant acceleration is $y(t) = v_0 t + \frac{1}{2} a t^2$, where $a$ is the acceleration, and $v_0$ is the initial velocity along displacement. $\endgroup$ – Nominal Animal Apr 2 '17 at 10:21
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    $\begingroup$ @ChubbyChoc Please see Narasimham's answer. Apparently, the question meant the amount of distance covered between $t=2$ and $t=3$, not the total distance. $\endgroup$ – projectilemotion Apr 2 '17 at 10:45
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Your teacher is correct. Question asks how much distance is covered between $t= 2$ and $ t= 3.$ Time lapse is 1 second, that is, in the third second of duration. In meters, distance travelled =

$$ s = \frac12 \cdot 10\cdot (3^2-2^2) = 25, $$

and, if you draw the parabola graph, $s_2-s_1 = a (t_2^2-t_1^2)/2. $

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    $\begingroup$ +1 and a very cheeky linguistic twist on the part of the test. I'd think it would be definitely grounds for a complaint to the administration. $\endgroup$ – James Snell Apr 2 '17 at 12:11
  • $\begingroup$ Ohhhhhhhh thanks! I didn't realise their use of language $\endgroup$ – ChubbyChoc Apr 2 '17 at 12:58
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    $\begingroup$ @JamesSnell No use methinks. English is the official language here. $\endgroup$ – Narasimham Apr 2 '17 at 14:00
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    $\begingroup$ It's not the choice of language, so much as their choice of wording. Even as a native speaker I know plenty of people who would have been caught out on a linguistic point, not a mathematical one. Looking at other answers here goes to illustrate that point. $\endgroup$ – James Snell Apr 3 '17 at 8:30
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You are correct. If the question is exactly how you phrased it, the displacement should be 45m down.

We know that:

a (acceleration) = $10m/s^2$

t (time) = 3 seconds

u (initial velocity) = $0m/s$

Hence, using the formula $s=ut-(1/2)at^2$ the answer should be 45 m.

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  • $\begingroup$ "In the 3rd second", not "in 3 seconds". $\endgroup$ – chepner Apr 2 '17 at 14:37
  • $\begingroup$ @chepner Sorry. I answered his question after he edited it. $\endgroup$ – David Hu Apr 3 '17 at 0:07
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We have

$$h=\frac {1}{2}gt^2+v_0t+h_0$$

$$=\frac {1}{2}gt^2$$

$$=\frac {1}{2}.10. (3)^2=45 m $$.

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  • $\begingroup$ That does not calculate displacement... $\endgroup$ – ChubbyChoc Apr 2 '17 at 10:01
  • $\begingroup$ @ChubbyChoc: $h(t) = \frac{1}{2} g t^2 + v_0 t + h_0$ most definitely does calculate displacement -- displacement at constant acceleration $g$ with initial velocity $v_0$ and initial displacement $h_0$. When initial velocity and displacement are zero, $h(t) = \frac{1}{2} g t^2$. $\endgroup$ – Nominal Animal Apr 2 '17 at 10:16
  • $\begingroup$ @NominalAnimal Yes, his/her comment is now obsolete because the question was changed. The OP answered the question before the change, and then changed his answer. $\endgroup$ – projectilemotion Apr 2 '17 at 10:19
  • $\begingroup$ @NominalAnimal oh ok my mistake $\endgroup$ – ChubbyChoc Apr 2 '17 at 10:24
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    $\begingroup$ @projectilemotion: Arrgh, I so hate these edit games. Thanks for pointing this out for me. For all others reading these comments, checking the original version of this answer shows that Salahamam_ Fatima originally had $h(t) = g t + v(0)$, for displacement with constant velocity. $\endgroup$ – Nominal Animal Apr 2 '17 at 10:26
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Using the formula $s=ut+1/2at^2$

Where $a$ is acceleration, $u$ is the initial velocity, $t$ is the time and $s$ the the displacement.

We can deduce that the displacement will be $45m$.

You are indeed correct!

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  • $\begingroup$ Did not see the edit haha. My bad. I answered the original question. $\endgroup$ – Sophie Quach Apr 2 '17 at 11:04

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