RUNWAY LENGTH GOAL Apply kinematics to horizontal motion with two phases. PROBLEM A typical jetliner lands at a speed of 1.60 × 10² mi/h and decelerates at the rate of (10.0 mi/h)/s. If the plane travels at a constant speed of 1.60 × 10² for 1.00s after landing before applying the brakes, what is

Question

RUNWAY LENGTH

GOAL Apply kinematics to horizontal motion with two phases.

PROBLEM A typical jetliner lands at a speed of [latex]1.60 imes 10^{2} \mathrm{mi} / \mathrm{h}[/latex] and decelerates at the rate of [latex](10.0 \mathrm{mi} / \mathrm{h}) / \mathrm{s}[/latex]. If the plane travels at a constant speed of [latex]1.60 imes 10^{2} \mathrm{mi} / \mathrm{h}[/latex] for [latex]1.00 \mathrm{~s}[/latex] after landing before applying the brakes, what is the total displacement of the aircraft between touchdown on the runway and coming to rest?

STRATEGY See Figure 2.18. First, convert all quantities to SI units. The problem must be solved in two parts, or phases, corresponding to the initial coast after touchdown, followed by braking. Using the kinematic equations, find the displacement during each part and add the two displacements.

Accepted Answer

Convert units of speed and acceleration to SI:

[latex]\begin{aligned}&v_{0}=\left(1.60 imes 10^{2} \cancel{ ext{ mi }} / \cancel{\mathrm{h}} ight)\left(\frac{0.447 \mathrm{~m} / \mathrm{s}}{1.00 \cancel{\mathrm{mi}} / \cancel{\mathrm{h}}} ight)=71.5 \mathrm{~m} / \mathrm{s} \&a=(-10.0(\cancel{ ext{ mi }} / \cancel{\mathrm{h}}) / \mathrm{s})\left(\frac{0.447 \mathrm{~m} / \mathrm{s}}{1.00 \cancel{ ext{ mi }} / \cancel{\mathrm{h}}} ight)=-4.47 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}[/latex]

Taking [latex]a=0, v_{0}=71.5 \mathrm{~m} / \mathrm{s}[/latex], and [latex]t=1.00 \mathrm{~s}[/latex], find the displacement while the plane is coasting:

[latex]\Delta x_{ ext {coasting }}=v_{0} t+\frac{1}{2} a t^{2}=(71.5 \mathrm{~m} / \mathrm{s})(1.00 \mathrm{~s})+0=71.5 \mathrm{~m}[/latex]

Use the time-independent kinematic equation to find the displacement while the plane is braking.

[latex]v^{2}=v_{0}{ }^{2}+2 a \Delta x_{ ext {braking }}[/latex]

Take [latex]a=-4.47 \mathrm{~m} / \mathrm{s}^{2}[/latex] and [latex]v_{0}=71.5 \mathrm{~m} / \mathrm{s}[/latex]. The negative sign on [latex]a[/latex] means that the plane is slowing down.

[latex]\Delta x_{ ext {braking }}=\frac{v^{2}-v_{0}{ }^{2}}{2 a}=\frac{0-(71.5 \mathrm{~m} / \mathrm{s})^{2}}{2.00\left(-4.47 \mathrm{~m} / \mathrm{s}^{2} ight)}=572 \mathrm{~m}[/latex]

Sum the two results to find the total displacement:

[latex]\Delta x_{ ext {coasting }}+\Delta x_{ ext {braking }}=71.5 \mathrm{~m}+572 \mathrm{~m}=644 \mathrm{~m}[/latex]

REMARKS To find the displacement while braking, we could have used the two kinematics equations involving time, namely, [latex]\Delta x=v_{0} t+\frac{1}{2} a t^{2}[/latex] and [latex]v=v_{0}+a t[/latex], but because we weren't interested in time, the time-independent equation was easier to use.

Question 2.6: RUNWAY LENGTH GOAL Apply kinematics to horizontal motion wit...

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