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PROBLEM STATES:

The landing velocity of an airplane (at which it touches the ground) is 100mi/hr. It decelerates at a constant rate and comes to a stop after traveling .25miles along a straight landing strip. Find the deceleration or the negative acceleration.

MY CONCERN/QUESTIONS:

I have worked to long on this problem and just can't find out how to approach this. I feel like i'm missing some important data or physical law that I need to apply. Usually we take the anti-derivative of acceleration make it equal the velocity in working backwards. However, I cannot approach this problem like this. I do not have acceleration and only have velocity.

To define initial conditions I would say at time (0) distance is (0). However, for some intuitive reason I'm wanting to say at time (0) distance is (.25).

Can anyone help me out in solving this problem. I feel like I'm not given a sufficient amount of information to solve this problem.

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4 Answers 4

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Let the deceleration be the positive number $a$. So the acceleration is $-a$.

Let $x=x(t)$ be the distance travelled from time of touchdown to time $t$.

Then $v=\frac{dx}{dt}$ is the velocity at time $t$, and $\frac{dv}{dt}=\frac{d^2 x}{dt^2}$ is the distance travelled on the ground by time $t$.

We have $\frac{dv}{dt}=-a$. Integrating, we find that $v=-at+C$ for some $C$. The velocity at time $0$ is $100$. So $C=100$.

Thus $\frac{dx}{dt}=-at+100$. Integrate. We get $x=-\frac{1}{2}t^2+100t+D$. But $x=0$ when $t=0$, so $$x=-\frac{1}{2}at^2+100t.\tag{$1$}$$

The velocity reaches $0$ at time $t_0$, where $-at_0+100=0$. So $t_0=\frac{100}{a}$.

Substitute this value of $t_0$ for $t$ in $(1)$, recalling the fact that at time $t_0$ we have $x=0.25$. So $$0.25=-\frac{1}{2}a\left(\frac{100}{a}\right)^2+100\left(\frac{100}{a}\right),$$ and solve for $a$.

Remark: The following is an easy approach not using integration. The rate of change of velocity is constant, so the average speed is $\frac{100}{2}$. We covered $0.25$ miles, so the time it took is $\frac{.25}{50}$ hours. Thus the rate of change of velocity was the negative of $\dfrac{100}{\frac{0.25}{50}}$. Thus the deceleration was $20000$ m/h$^2$.

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  • $\begingroup$ This is massively over-complicated. The OP just needs to apply the formula $v^2 = u^2 + 2as$. $\endgroup$ Commented Apr 9, 2013 at 1:43
  • $\begingroup$ See here: ronknott.com/MEI/MechSuvatEquns.html $\endgroup$ Commented Apr 9, 2013 at 1:45
  • $\begingroup$ That sums it up quite nice. I see where I have been going wrong. Thanks $\endgroup$ Commented Apr 9, 2013 at 1:45
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Time of deceleration $T$

$$v_0 + a T = v(T) = 0 \implies a = -\frac{v_0}{T}$$

where $v_0$ is the initial speed, $a$ is the deceleration. The distance traveled $d$ during this deceleration is:

$$d = v_0 T + \frac{1}{2} a T^2$$

Know $d$, $v_0$, seeking $a$, $T$.

$$d = v_0 T + \frac{1}{2} \left (-\frac{v_0}{T} \right ) T^2 = \frac{1}{2} v_0 T \implies T=\frac{2 d}{v_0}$$

Plug back into first equation

$$a = -\frac{v_0}{T} = -\frac{v_0^2}{2 d}$$

Plug in $v_0 = 100$ mph and $d = 0.25$ mi:

$$a = -\frac{10,000}{0.5} = -20,000 \:\text{miles per hour}^2$$

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  • $\begingroup$ I'm having a rough time understanding what you said. Could you please clarify . $\endgroup$ Commented Apr 9, 2013 at 1:19
  • $\begingroup$ @ShaneYost: I made some edits. What exactly are you having trouble understanding? $\endgroup$ Commented Apr 9, 2013 at 1:20
  • $\begingroup$ I tried using this approach if i'm understanding it correctly where I solve for T first. But when I do I end up with a value slightly over 21,000. The answer in the book says it's 20,000. Therefore, i cannot get my numbers to balance out resulting in 20,000mi/hr^2 $\endgroup$ Commented Apr 9, 2013 at 1:24
  • $\begingroup$ @ShaneYost: You have 2 equations and 2 unknowns. In the first, I solved for $a$ in terms of $T$. In the second, I plugged in that solution and got a simple equation for $T$. I then solved for $T$, and then plugged that solution into the equation for $a$. $\endgroup$ Commented Apr 9, 2013 at 1:27
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We can solve this problem given $a = \frac{dv}{dt}$

Split the fraction:

$$adt = dv$$

Integrate both sides, given that acceleration is a constant:

$$\int a dt = \int dv$$ $$a \int dt = \int dv$$ $$at = v_f - v_0$$ $$v_f = v_0 + at$$

Now we know that $v = \frac{dx}{dt}$ so we can substitute that and integrate again:

$$\frac{dx}{dt} = v_0 + at$$ $$dx = (v_0 + at)dt$$ $$\int dx = v_0\int dt + a\int t dt$$

Which gives us our working equation:

$$x_f - x_0 = v_0 t + \frac{1}{2}at^2$$

Note that $x_f - x_0$ is just d, your distance. We get that $t = \frac{v_f - v_0}{a}$ (and just to simplify since the plane comes to a complete stop $v_f = 0$ and $t = \frac{-v_0}{a}$ so we can plug in t in our working equation:

$$d = -\frac{v_0^2}{2a}$$ $$a = -\frac{v_0^2}{2d}$$

From here you can plug in your values.

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This is a standard mechanics question. We apply the SUVAT equations of motion. You have the initial velocity $u=+100$ mph, you have the final velocity $v= 0$ mph, you have the displacement $s = 0.25$ miles and you want the acceleration $a=?$ m/h/h. These are connected by:

$$v^2 = u^2 + 2as$$

Substituting in our values: $0^2 = 100^2 + 2 \times a \times 0.25$ which gives $0 = 10000 + \tfrac{1}{2}a$. Solving for $a$ gives $a = -20,000$ miles per hour, per hour.

I know that that sounds massive, but $100$ miles per hour is about $45$ metres per second and $0.25$ miles is about $400$ metres. Using the correct figures gives $a\approx -2.5 \, \text{ms}^{-2}$.

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