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Chapter 2: Problem 34

An airplane lands and starts down the runway with a southwest velocity of $55 \mathrm{m} / \mathrm{s}$. What constant acceleration allows it to come to a stop in \(1.0 \mathrm{km} ?\)

Short Answer

Expert verified
Answer: The constant acceleration required for the airplane to come to a stop in 1.0 km is -1.51 m/s虏.

Step by step solution

01

List the known quantities

We know the following: - Initial Velocity \((v_i)\): \(55~m/s\) (southwest) - Final Velocity \((v_f)\): \(0~m/s\) (since the airplane comes to a stop) - Displacement \((d)\): \(1000~m\) (1.0 km)
02

Choose the appropriate equation of motion

We can choose the equation of motion that relates initial velocity, final velocity, acceleration, and displacement: \(v_f^2 = v_i^2 + 2ad\)
03

Solve for acceleration

Using the equation of motion from Step 2, we can plug in the known values and solve for the acceleration (\(a\)): \((0~m/s)^2 = (55~m/s)^2 + 2ad\) \(0 = 55^2 + 2ad\) Now, we'll isolate \(a\) on one side of the equation: \(-55^2 = 2ad\) Now divide both sides by \(2d\): \( a = \frac{-55^2}{2(1000)}\)
04

Calculate the acceleration

Calculate the acceleration (\(a\)) using the numbers in the equation: \( a = \frac{-55^2}{2(1000)}\) \( a = \frac{-3025}{2000} \approx -1.51~m/s^2\) The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which is the case because the airplane is decelerating to come to a stop. Thus, the constant acceleration required for the airplane to come to a stop in 1.0 km is \(-1.51~m/s^2\).

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