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`        an express train moving at 30m/s reduses its speed to 10m/s in a distance of 240m if the breaking force is increased by 12.5% in the begining find the distane it that travels before coming to rest`
4 years ago

deeksh bharath
11 Points
```							given that the train is moving initially at 30m/s and then decelerates to 10m/s thenv2_u2=2as(10)2-(30)2=2(-a)(240)100-900=(-480a)-800= -480aa= -5/3m/s2again tis given that if initially the breaking force is 12.5%the by f=ma, as f increases ,aceleration also increases,thereforenew aceleration is a( 1+12.5/100 )(-50/3)a=1.875 or 15/8m/s2now use the sformula u2/2a             d = (30)2/2*(-15/8)m               d = 240mtherefore the train will come to rest or stom at the ditance of 240m if the breaking force is 12.5% in the initial.
```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions