Question

Please solve for acceleration in the same way that position and velocity are solved with. Using theta, theta double dot etc

SOUTH TEXAS COLLEGE ENGR2302-Dynamics Dynamics Project - Velocity & Acceleration Derivation Derive the Velocity and Acceleration of the piston from its position equation. The velocity has been derived, look at the video. Submit a detail derivation of the Acceleration. Note: you will use that derivation for your final project, so it will determine how many points you will get on your project. 02 0.75 Ft S = x S = 0.2 cose + 0.752 - (0.2sino) -1/ = V=-0.2 singø +(0.75= (0,29166).-2(012 Sin6). 9.2. cos 6€ dt dv ot -9-a Taking the derivative of (inside)2. Drop the exponent, the inside stays the same, subtract 1 from the exponent, and times take the derivative of the inside.

Answer 1

8= 0.2 C08 @ +6045)2 = (0-2 Siuo)? Velocity, v= ds at v=d dt (0:2.Com td (0-75)2 – (0.25iuo)? dt Х 2 V=-0.2 Stub + 110799 –10:2Souse) 7-12 fo- 2 (0-2 siu) (0-2 lede) Х N=-0.2 (Sind a -0.04 Sind coso o 10.75) (0.2 Sius) (*** (0.2 Stua)?]*2 Now, fideig Aralisation a do -2(6 * 18.40 ) Bouleuit Band it fosso colo dit die blog

Illo 75)2 – (0.2 Siue) X Х 0-210:2) Suo (0-2 Cose -0.04 Sino Coso o 0.04 Siu o cosd Q = -0.2 sind ö + Côso 03) -0.04 Sin 0.com +(coso :)? - (Scud on (1013 (0°75)2 – (0:2 Sino) 1/2 a=_-0-2 (Sus öt csille os ?) = (studemo ö + loro 0) ?-tsium Sino cose & (6.75) - (0.9 Sino? -3/2 103) Autwel