/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 An aeroplane is rising verticall... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 35

An aeroplane is rising vertically with acceleration \(f\). Two stones are dropped from it at an interval of time \(t\). The distance between them at time \(t^{\prime}\) after the second stone is dropped will be (A) \(\frac{1}{2}(g+f) t t^{\prime}\) (B) \(\frac{1}{2}(g+f)\left(t+2 t^{\prime}\right) t\) (C) \(\frac{1}{2}(g+f)\left(t-t^{\prime}\right)^{2}\) (D) \(\frac{1}{2}(g+f)\left(t+t^{\prime}\right)^{2}\)

Short Answer

Expert verified
The short answer based on the step-by-step solution is: The distance between the two stones at time \(t'\) after the second stone is dropped is: \(s = \frac{1}{2}(g+f)(t')^2\left( (t'+t)^2 - 1 \right)\) None of the given options match this result.

Step by step solution

01

Analyze the motion of the first stone

The first stone is dropped at time \(t_1 = 0\). When it is dropped, it has the same upward acceleration as the airplane, which is \(f\). After the stone is dropped, it is under the influence of gravity, so its acceleration changes to \(g\). Thus, its net acceleration is \(a_1 = g + f\). Now let's find the position of the first stone at any time \(T\). We can use the equation of motion: \(s = ut + \frac{1}{2}at^2\). Since the stone is dropped, its initial velocity \(u\) is zero. Therefore the equation becomes: \(s_1 = \frac{1}{2}(g+f)T^2\).
02

Analyze the motion of the second stone

The second stone is dropped at time \(t_2 = t\). Similarly, when it is dropped, it has the same upward acceleration as the airplane \((f)\) and experiences gravity \(g\) after being dropped. Thus, its net acceleration is \(a_2 = g + f\). We need to find its position at time \(t'\) after it is dropped, which corresponds to time \(T = t + t'\) for the second stone. Again, we can use the equation of motion: \(s = ut + \frac{1}{2}at^2\). Since the stone is dropped, its initial velocity \(u\) is zero. So the equation becomes: \(s_2 = \frac{1}{2}(g+f)(t+t')^2\).
03

Calculate the distance between the two stones

The distance between the two stones at time \(T = t + t'\) can be found by subtracting their positions as calculated in step 1 and step 2. Distance between stones = \(s_2 鈥� s_1\) \(s = \frac{1}{2}(g+f)(t+t')^2 - \frac{1}{2}(g+f)T^2\) Substitute \(T = t + t'\) in the equation: \(s = \frac{1}{2}(g+f)(t+t')^2 - \frac{1}{2}(g+f)(t+t')^2\) Simplifying this equation, we get: \(s = \frac{1}{2}(g+f)(t'+t)^2 - \frac{1}{2}(g+f)(t')^2\) Factor out \(\frac{1}{2}(g+f)(t')^2\): \(s = \frac{1}{2}(g+f)(t')^2\left( (t'+t)^2 - 1 \right)\) Comparing with given options, we find that the answer is closest to option (D) but is not exactly matching. Therefore, we can conclude that none of the given options are correct.

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