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A particle is moving in x-y plane at 2m/s along the x-axis, 2 seconds later, it’s velocity is 4m/s in a direction making 60 o with the positive x-axis. It’s average accelaration for this period of motion is: (a) root 5 m/s 2 , along y-axis (b) root 3 m/s 2 , along y-axis (c) root 5 m/s 2 , along at 60 o with the positive x-axis (d) 3 m/s 2 , at 60 o with the positive x-axis


4 years ago

RANJEET KUMAR
32 Points
							To solve this question we need the basic understanding of rectangular coordinate axes and how to break a vector into its rectangular components.lets take +ve x axis to be $\widehat{i}$ and +ve ‘y’ axis to be $\widehat{j}$ .so initial velocity of the particle  = 2m/s $\widehat{i}$, and final velocity of particle after breaking it into its coordinates =  2$\widehat{i}$ + 2$\sqrt[\sqrt[]{3}$$\widehat{j}$ in m/s.so average acceleration = [final velocity – initial velocity]/total time taken                                       =[[2$\widehat{i}$ + 2$\sqrt[\sqrt[]{3}$$\widehat{j}$] – 2$\widehat{i}$] / 2                                       =$\sqrt[\sqrt[]{3}$$\widehat{j}$ m/s2hence the correct answer will be option [b]thanku very much!

4 years ago
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### Course Features

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