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solution
convert units of speed and acceleration to si.
$v_0=(1.60\times 10^{2} mi/h)(\frac{0.447 m/s}{1.00 mi/h}) = 71.5 m/s$
$a=(-10.0 (mi/h)/s)(\frac{0.447 m/s}{1.00 mi/h})=-4.47 m/s^{2}$
taking $a = 0$, $v_0=71.5 m/s$, and $t = 1.00 s$, find the displacement while the plane is coasting.
$delta x_{coasting}=v_0t+\frac{1}{2}at^{2}=(71.5 m/s)(1.00 s)+0 = 71.5 m$
use the time - independent kinematic equation to find the displacement while the plane is braking.
$v^{2}=v_0^{2}+2adelta x_{braking}$
take $a=-4.47 m/s^{2}$ and $v_0 = 71.5 m/s$. the negative sign on $a$ means that the plane is slowing down.
$delta x_{braking}=\frac{v^{2}-v_0^{2}}{2a}=\frac{0-(71.5 m/s)^{2}}{2.00(-4.47 m/s^{2})}=572 m$
sum the two results to find the total displacement.
$delta x_{coasting}+delta x_{braking}=71.5 m + 572 m=644 m$
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remarks to find the displacement while braking, we could have used the two kinematics equations involving time, namely, $delta x=v_0t+\frac{1}{2}at^{2}$ and $v = v_0+at$, but because we werent interested in time, the time - independent equation was easier to use.
question by how much would the answer change if the plane coasted for 2.0 s before the pilot applied the brakes?
enter a number m
Step1: Calculate new coast - ing displacement
When $v_0 = 71.5$ m/s, $a = 0$ m/s² and $t = 2.0$ s, use the formula $\Delta x=v_0t+\frac{1}{2}at^{2}$. Since $a = 0$, $\Delta x_{coasting}=(71.5\ m/s)\times(2.0\ s)+0 = 143$ m.
Step2: Braking displacement remains the same
The braking displacement $\Delta x_{braking}$ is still calculated using $v^{2}=v_{0}^{2}+2a\Delta x_{braking}$. With $v = 0$, $v_0 = 71.5$ m/s and $a=-4.47$ m/s², $\Delta x_{braking}=\frac{v^{2}-v_{0}^{2}}{2a}=\frac{0-(71.5\ m/s)^{2}}{2\times(-4.47\ m/s^{2})}=572$ m.
Step3: Calculate new total displacement
The new total displacement $\Delta x_{total}=\Delta x_{coasting}+\Delta x_{braking}=143\ m + 572\ m=715$ m.
Step4: Calculate the change in displacement
The original total displacement was 644 m. The change $\Delta=\Delta x_{total}-644\ m=715\ m - 644\ m = 71$ m.
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