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        A particle moves along the parabolic path Y equal to ax2 in such a way that the X component of the velocity remains constant say c. The acceleration of the particle is1. ackcap2. 2ac2jcap3.2ac2kcap4.a2cjcap
one year ago

Khimraj
3008 Points
							acceleration  = ax$\hat{i}$ + ay$\hat{j}$= (dvx/dt)$\hat{i}$ + (dvy/dt)$\hat{j}$ = (dc/dt)$\hat{i}$ + (d2y/dt2)$\hat{j}$= 0 + d/dt[2ax(dx/dt)]$\hat{j}$= d/dt[2acx]$\hat{j}$= 2ac2 $\hat{j}$Hope it clears. If you like answer then please approve it.

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