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Chapter 2: Problem 36

A 1200 -kg airplane starts from rest and moves forward with a constant acceleration of magnitude \(5.00 \mathrm{m} / \mathrm{s}^{2}\) along a runway that is 250 m long. (a) How long does it take the plane to reach a speed of \(46.0 \mathrm{m} / \mathrm{s} ?\) (b) How far along the runway has the plane moved when it reaches \(46.0 \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
What is the distance traveled along the runway when the airplane reaches this speed? Answer: The airplane takes 9.20 seconds to reach a speed of 46.0 m/s, and when it reaches this speed, it has moved 211.6 meters along the runway.

Step by step solution

01

Finding the time to reach a speed of 46.0 m/s

To find the time to reach a speed of 46.0 m/s, we will use the kinematic equation: v = u + at Here, v = 46.0 m/s (final velocity), u = 0 m/s (initial velocity, as the airplane starts from rest), and a = 5.00 m/s^2 (acceleration). We need to find t (time). Rearranging the equation to solve for t, we get: t = (v - u) / a Substituting the given values: t = (46.0 - 0) / 5.00 t = 46.0 / 5.00 t = 9.20 s So, it takes the plane 9.20 seconds to reach a speed of 46.0 m/s.
02

Finding the distance travelled along the runway

Now, we'll use the kinematic equation to find the distance travelled along the runway when the plane reaches 46.0 m/s: s = ut + (1/2)at^2 Here, s is the distance, u = 0 m/s (initial velocity), t = 9.20 s (time, which we calculated in Step 1), and a = 5.00 m/s^2 (acceleration). Substituting the given values: s = 0 * 9.20 + (1/2) * 5.00 * 9.20^2 s = 2.5 * 9.20^2 s = 2.5 * 84.64 s = 211.6 m So, when the plane reaches a speed of 46.0 m/s, it has moved 211.6 meters along the runway.

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