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Introductory Physics Homework Help
Find the value of ##t## when ##P## returns to ##X##- Kinematics
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[QUOTE="chwala, post: 6851282, member: 287397"] [USER=47950]@sammy[/USER], the distance of ##P## from ##X## when the particle is at ##t=12## seconds is equal to ##3.6## metres. Unless, you also want me to show this? Ok, let me take you through this; 1. The distance of ##P## from ##X## when ##t=6## is given by; ##s=\dfrac{1}{2} ×6 × 2.4= 7.2## metres 2. The distance of ##P## from ##X## when ##t=12## is given by; i.e from the point where ##t=6## with initial velocity ##u=0##. ##v^2=u^2+ 2as## ##3.6^2=0^2+ (2 ×0.6×s)## ##12.96=0+ (2 ×0.6×s)## ##s=10.8## metres Therefore, the required ##s## value is given by; ##s_{required}=10.8-7.2=3.6## metres. I hope its no longer a mystery :wink: ... Cheers. [/QUOTE]
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Introductory Physics Homework Help
Find the value of ##t## when ##P## returns to ##X##- Kinematics
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