Question

question 25
1 pts
a box has an initial velocity of \v\ m/s. it moves across a rough, flat surface and eventually comes to a stop after traveling a distance of \d\ meters. if the velocity is doubled to \2v\ what will be the new distance it moves across the ground before stopping in terms of the original distance \d\?

○ d/4
○ 2d
○ d
○ d/2
○ 4d

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Explanation:

Step1: Use the kinematic equation

The kinematic equation \(v^{2}=u^{2}+2as\). When the box stops, \(v = 0\). The initial velocity is \(u\), acceleration is \(a\) (negative as it is decelerating) and distance is \(s\). So, \(0 = u^{2}+2as\), which gives \(s=\frac{-u^{2}}{2a}\).

Step2: Analyze the first - case

In the first case, \(u = v\) and \(s = d\). So, \(d=\frac{-v^{2}}{2a}\).

Step3: Analyze the second - case

In the second case, \(u = 2v\). Let the new distance be \(s_{new}\). Then \(s_{new}=\frac{-(2v)^{2}}{2a}\).

Step4: Substitute \(a\) from the first - case into the second - case

From \(d=\frac{-v^{2}}{2a}\), we can get \(a=\frac{-v^{2}}{2d}\). Substitute \(a\) into \(s_{new}=\frac{-(2v)^{2}}{2a}\).

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Answer:

\(4d\)