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.
Best Answer
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$^2$.