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An aircraft is flying at a speed of $200 \text{m/s}$. As it is coming in to land the aircraft reduces its speed from $200\text{m/s}$ to $50\text{m/s}$ at a constant rate of $2.5\text{m/s$^2$}$. How many metres does it travel in this time?

So what I did was $200 - 50 = 150$ and $150 / 2.5 = 60\text{s}$, and calculated (meters)

$$\sum_{k=1}^{60}200-2.5k$$

The result of it was $12000 - 4575 = 7425$, however the answer is $7500\text{m}$. I don't know where I went wrong.

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    Are the formulas $v=v_0+at,e=e_0+v_0t+at^2/2$ unavailable ?? – 2017-02-21
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    You are wrong to believe that a discrete summation yields the correct answer. – 2017-02-21
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    This question deserves to be asked at physics.stackexchange – 2017-02-21

8 Answers 8

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Hint:

The answer can be obtained directly by just using the formula $$v_{\text {final}}^2-v_{\text{initial}}^2 =2as $$ where $s $ is the distance travelled. Hope it helps.

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    That is a nice formula, but what is 'a' then? – 2017-02-21
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    @Helena $a $ is the acceleration. – 2017-02-21
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Let us play your game with seconds:

$$e=\sum_{k=0}^{59}\left(200-2.5k\right)=200\cdot60-2.5\frac{59\cdot60}{2}=7575$$

(this is different from your answer because I am summing from $0$ to $59$).

Then using tenths of a second instead, updating all parameters accordingly:

$$e=\sum_{k=0}^{599}\left(20-0.025k\right)=20\cdot600-0.025\frac{599\cdot600}{2}=7507.5$$ And now with milliseconds:

$$e=\sum_{k=0}^{59999}\left(0.2-0.0000025k\right)=0.2\cdot60000-0.0000025\frac{59999\cdot60000}{2}=7500.075$$

With microseconds, you would obtain $7500.000075$, so where's the truth ?

You can infer that with smaller and smaller units of time, you get a better and better approximation, and eventually the right answer, $7500$.

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$$ \begin{align} {\small\text{initial conditions: }}\quad \,v_{\small0}\,\, &= 200\, \,{\small\text{ at }}\, \,t_{\small0}=0\, \,{\small\text{ where }}\, \,s_{\small0}=0\, \\[2mm] \frac{d^2s}{dt^2} &= \frac{dv}{dt} = a=\color{red}{-2.5}=-\frac52 \\[2mm] {\small\text{Integrate once: }}\quad \frac{ds}{dt}\, &= v=-\frac52t+C_{\small1} \\[2mm] {\small\text{Apply initial: }}\quad\,\, v_{\small0}\,\, &= -\frac52t_{\small0}+C_{\small1} \Rightarrow C_{\small1}=200 \Rightarrow \frac{ds}{dt}=\color{red}{v=-\frac52t+200}\tag{1} \\[2mm] {\small\text{Integrate again: }}\quad\,\,\, s\,\,\, &= -\frac54t^2+200t+C_{\small2} \\[2mm] {\small\text{Apply initial: }}\quad\,\, s_{\small0}\,\, &= -\frac54t_{\small0}^2+200t_{\small0}+C_{\small2} \Rightarrow C_{\small2}=0 \Rightarrow \color{red}{s=-\frac54t^2+200t}\tag{2}\\[2mm] \end{align} $$
Use $(1)$ and $(2)$ to calc the system. For instance,
As of the question: $$ v=50 \,\Rightarrow\, t=\frac{v-200}{-5/2}=60 \,\Rightarrow\, s=-\frac54t^2+200t=7500 $$ When it stops: $$ v=0 \,\Rightarrow\, t=\frac{v-200}{-5/2}=80 \,\Rightarrow\, s=-\frac54t^2+200t=8000 $$ Assuming minus velocity is backward movement; it should return-back with-in: $$ s=0 \,\Rightarrow\, -\frac54t^2+200t=0 \rightarrow t=160 \,\Rightarrow\, v=-\frac52t+200=-200 $$

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$$ s = \dfrac{v^2-u^2}{2 a} = \dfrac{250\cdot 150}{2\cdot 2.5} = 7500 $$

Integration with infinitely small intervals is reckoned in calculus.

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$$S=ut+1/2at^2$$

$$=200 \times 60 + 1/2 \times-2.5\times60^2$$ $$=12000-4500$$ $$=7500$$

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Let $travel_m = as^2+bs+c$ (s = time, travel_m = travelled distance)

$\frac{\text{d}^2y}{\text{d}s^2} as^2+bs+c$ (The second derivative of distance gives you the acceleration.)

$=2a=-2.5$

$a=-1.25$

$travel_m = -1.25s^2+bs+c$

At first, the travel distance was $0$, $c=0$.

$travel_m = -1.25s^2+bs$

It's speed was $200$, and the time, $s$, was zero.

$\frac{\text{d}y}{\text{d}s}-1.25s^2+bs$ (The first derivative of distance gives you the speed.)

$= -2.5s + b=200$ ($s=0$)

$b = 200$

$travel_m = -1.25s^2 + 200s$

When is the speed 50 m/s? Let's find.

$\frac{\text{d}y}{\text{d}s}-1.25s^2+200s$

$= -2.5s +200 = 50$

$= -2.5s = -150$

$ s = 60$

$-1.25s^2 + 200s$, substitute $s=60$

$= 7500$

  • Graph of travelled distance:

  • Graph of speed

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Hint: as long as the speed varies linearly (i.e. with constant acceleration), the distance traveled during an interval of time $t$ is simply $\bar v \cdot t$ where $\bar v$ is the average of the beginning and ending speeds. In this case $\,t=(200-50)/2.5\,$ and $\,\bar v = (200+50)/2\,$.

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We have $$ v(t_1) = 200 \, \frac{\text{m}}{\text{s}} \\ v(t_2) = 50 \, \frac{\text{m}}{\text{s}} $$ and an acceleration of $$ \dot{v}(t) = -2.5 \, \frac{\text{m}}{\text{s}^2} $$ Integrating both sides gives: $$ \int\limits_{t_1}^{t_2} \dot{v}(t) \, dt = -2.5 \, \frac{\text{m}}{\text{s}^2} \int \limits_{t_1}^{t_2} dt \iff \\ v(t_2) - v(t_1) = - 2.5 \, \frac{\text{m}}{\text{s}^2} (t_2-t_1) $$ so we have $$ t_2-t_1 = \frac{50 \, \frac{\text{m}}{\text{s}} -200 \, \frac{\text{m}}{\text{s}} }{(- 2.5 \, \frac{\text{m}}{\text{s}^2})} = 60 \, \text{s} $$ Integrating $$ \dot{s}(t) = 200 \, \frac{\text{m}}{\text{s}} - 2.5 \, \frac{\text{m}}{\text{s}^2} (t-t_1) $$ gives \begin{align} s(t_2) - s(t_1) &= 200 \, \frac{\text{m}}{\text{s}} (t_2 - t_1) - \frac{5}{4} \, \frac{\text{m}}{\text{s}^2} (t_2-t_1)^2 \\ &= 200 \, \frac{\text{m}}{\text{s}} 60 \, \text{s} - \frac{5}{4} \, \frac{\text{m}}{\text{s}^2} (60 \, \text{s})^2 \\ &= 7500\, \text{m} \end{align}