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`        A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3 -2t), where t is in second and velocity is in m/s.What is the acceleration of the particle, when it is at distance 2m from the origin. `
2 years ago

Pranav singh
18 Points
```							V=4t^3-2ta=dv/dt=12t^2-2.......(1)ds/dt=v=4t^3 - 2tds=(4t^3-2t)dtIntegrate both side.. Limit of s is 0 to 2 and t is 0 to t2=t^4-t^2 t^4-2t^2+t^2-2=0(t^2 +1)(t^2 - 2)=0..neglecting t=-1t^2=2..a=12t^2 - 2a=12×2-2a=22m/s
```
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