{"id":23316,"date":"2026-01-14T16:00:56","date_gmt":"2026-01-14T21:00:56","guid":{"rendered":"https:\/\/www.purdue.edu\/freeform\/me274\/?p=23316"},"modified":"2026-01-17T15:53:21","modified_gmt":"2026-01-17T20:53:21","slug":"homework-h1-d-sp-26","status":"publish","type":"post","link":"https:\/\/www.purdue.edu\/freeform\/me274\/2026\/01\/14\/homework-h1-d-sp-26\/","title":{"rendered":"Homework H1.D – Sp 26"},"content":{"rendered":"\n\n\n
Problem statement<\/strong><\/em><\/span><\/a>
\nSolution video<\/a>
\n<\/strong><\/em><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\n

DISCUSSION THREAD<\/span><\/strong><\/em><\/p>\n

\"\"<\/p>\n

Discussion and hints:<\/strong><\/em><\/p>\n

Here, the distance traveled along the circular path, s<\/em>, is known as an explicit function of time. The speed of P at any instant is simply v = ds\/dt, and the rate of change of speed is: v_dot = d2<\/sup>s\/dt2<\/sup>. Since the path is circular, the radius of curvature is given by: \u03c1 = R<\/em>. With these three quantities, you have everything that you need to write down the velocity v<\/strong> = v e<\/strong>t\u00a0<\/sub><\/em>\u00a0and a<\/strong> = v_dot e<\/strong>t<\/sub>\u00a0+ (v2<\/sup>\/\u03c1) e<\/strong>n<\/sub>.<\/em><\/p>\n

Carefully watch the animation below of the motion of P. When the acceleration vector shown in RED<\/em><\/span> points “backward” from the direction of travel, particle P is slowing down; that is, the rate of change of speed is negative. Conversely, when the acceleration vector “forward” of the direction of travel, P is increasing in speed with a positive rate of change of speed. You will see that the period of time during which the speed of P is increasing is rather short, occurring during a time immediately after the direction of P changes.<\/p>\n

<\/p>\n

\n

Any questions??<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

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