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Answer :
Final answer:
To solve this physics problem, we use the equations of motion. Given the final velocity of 150 miles/hr (converted to m/s) and acceleration distance of 1200 m, we calculate the acceleration. Using this acceleration in the equation of velocity, we calculate the time it takes for the aircraft to reach take-off speed, approximately 35.8 seconds. The correct option is 1.
Explanation:
The question is a typical physics problem relating to motion. To solve this, you have to use the formula which connects distance (s), initial velocity (u), final velocity (v), and acceleration (a): v² = u² + 2as.
Given that the initial velocity (u) of the plane is 0 m/s (as it starts from rest), the final velocity (v) is 150 miles/hr (which is approximately 67.056 m/s when converted to m/s), and the distance over which the aircraft accelerated is 1200 m.
By substituting these values into the formula, we get (67.056 m/s)² = 0 + 2 * a * 1200 m. From this, we can calculate the acceleration (a) of the aircraft, which is approximately 1.877 m/s². Now, we apply this acceleration into the equation v = u + at to find the time (t) it takes for the aircraft to reach take-off speed.
When we plug in the values into this equation, we get 67.056 m/s = 0 + 1.877 m/s² * t. Solving this for t gives us t = 67.056 m/s / 1.877 m/s², which is approximately 35.71 seconds or rounded off to 35.8 seconds. So, the correct answer is (1) 35.8 seconds.
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