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Exploring Constant Acceleration in Physics: Class Discussion Topics

Join us for a lively discussion on constant acceleration in physics! We will delve into essential concepts, including free fall, reaction times, and the equations of motion. Students are invited to submit their questions through Google Moderator for consideration in our class on Friday. Whether you're curious about the equations of motion or specific scenarios like the rocket launch analysis, this is your chance to engage and clarify your understanding. Don't miss this opportunity to enhance your knowledge and participate in a hands-on learning experience!

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Exploring Constant Acceleration in Physics: Class Discussion Topics

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  1. PH 211 Winter 2014 Wednesday January 22

  2. Google moderator Which topic would you like to discuss in class on Friday? http://www.google.com/moderator/#16/e=20c5fe

  3. Google moderator Today’s questions: http://www.google.com/moderator/#16/e=211578

  4. Three dimensions.

  5. Three useful equations1D, CONSTANT ACCELERATION t v s

  6. Which of the following graphs represents motion with constant acceleration? A B t x x t v C D x t v

  7. Which of the following graphs represents motion with constant acceleration?PRESS ENTER TO SUBMIT MULTIPLE ANSERS! • A • B • C • D A B t x x t v C D x t v 0 of 208 :01

  8. Falling bodies starting at restThey all fall with the same acceleration if one ignore air resistance, remember moon video.

  9. Reaction time Median = 0.16 Mean = 3.48 Press 0.XX and ENTER to submit 0 of 208 :58

  10. A rocket is launched straight up with constant acceleration. 4.0s after liftoff, a bolt falls of the side of the rocket. The bolt hits the ground 6.0s later. What was the acceleration of the rocket?

  11. A rocket is launched straight up with constant acceleration. 4.0s after liftoff, a bolt falls of the side of the rocket. The bolt hits the ground 6.0s later. What was the acceleration of the rocket? • True • False I have an answer or I give up 137 of 208

  12. What is known. At t=0s we have s=0m, v=0m/s At t=4.0s we change a=aup to a=-g At t=10.0s we have s=0m What we need to find: aup.

  13. s t

  14. Upward motion: s(t)=1/2 aup t2 v(t)=aup t Downward motion s(t)=s(4)+v(4)(t-4)-1/2 g (t-4)2 v(t)=v(4)-g(t-4) End 0=s(10)=s(4)+6v(4)-18g

  15. Upward motion: s(t)=1/2 aup t2 v(t)=aup t End 0=s(10)=s(4)+6v(4)-18g 0= 8aup + 24 aup - 18g aup = 9g/16

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