A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation
Text Solution
Verified by ExpertsB
Let ' a' be the retardation of boggy then distance covered by it be S . If u is the initial velocity of boggy after detaching from train ( i.e. uniform speed of train)
\(v^{2} - u^{2} + 2\alpha ,\; 0 - u^{2} - 2\alpha ,\; s_{c} = - \frac{u^{2}}{2\alpha}\)
Time taken by boggy to stop
\(\nu - \nu + \alpha = 0 - \nu - \alpha \Rightarrow t = \frac{\nu}{\alpha}\)
In this time t distance travelled by train \(- \frac{1}{t} - \frac{q}{t} - \frac{q'}{c}\)
Hence ratio \(\frac{1}{2} = \frac{2}{4}\)
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems