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        position of particle moving along x axis is given by x=3t-4t^2+t^3 where x is in meter and t in second find average velocity of particle in time interval t=2 to t=4 sec
one year ago

Arun
23357 Points
							differentiate the equation w.r.t. time n put value of T. hope it helps RegardsArun (askIITians forum expert)

one year ago
Khimraj
3008 Points
							x = t3-4t2+3tv = dx/dt = 3t2 – 8t + 3average velocity = (1/T)$\int_{0}^{T}vdt$= $(1/2)\int_{2}^{4}(3t^{2}-8t+3)dt$= $(1/2)[(t^{3}-4t^{2}+3t)]_{2}^{4}$= 14/2 = 7m/sHope it clears.................

one year 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