Question

Σ A machine's motion is as shown below. NOTE: The line is NOT straight between t=6 and t=8s. 1) Draw the s-t graph and a-t graph. Label key points, axes; clearly distinguish between straight and curved lines. Enter the value of acceleration, a, at t=8 s, for the online answer below. 2) Determine the function of acceleration (a) and distance (s) in terms of time (t) over the time interval 6 to 8 seconds. 15 12 . V = 36-0.52 0 from the

Answer 1

First of all we'll try to get the equations corresponding to the velocity of motion.

Once we get the equations, then by using displacement velocity and velocity acceleration relationship , we'll get the equations of displacement as well as acceleration.

Although in the question, equations corresponding to t= 6s to t=8s only is asked, but I have provided you with equations of the complete motion, so that you can get displacement velocity and acceleration at any instant.

Calculations are given in the following images::

A ^ (6,18) (5,18) (maps) V=36–0.57 E + 7 (0,0) 7 (s) ย. (6-13) (PR'or) - equation of AB :- (V-0) = 62.09 )--) P v=-4.5t (2,, 9,) equation of Straight edi alt equation of Br in 'h-BR =('h-R) 15-x) nebo 18-1 t-2 2 [v-1-93] = 1/2 5-2 V+9 = gt V = gt-27 of con equation V=18 Vn- 4.5t 7 > 7>o Vogt-¿7 wat 2K tes V-18 for 54teG v 88779 = 36.0.54² for

y for Displacement in D for octaal- ~= -4.5t da dt Sare = fa.stat -45 + for octcr 2 qi85 +2+ c Boundary condition in :30, n=0 -0 + c Ceo x = -2.25te oct se u at t=es => e = -2.25legea .gm 6 for estas ir V gt-27 dr = Svet = Set-27)+ Vat x = 11 4.5+² - 27t to x Boundary conditionin for t=85 2- -9m 4.51€ ge- &&legte 9 ca 27 for 4.5+&- 27t+E7 x = estas at ter at ts35 5 second x= 4,515)-2765 )+22 –+4,5 m => > = 4,5739&2713)+27 = -13.5m o= 4+² -27ttet t=1.2*, 4.73 Х for n=0 docsmag. foue falls in estas at 25tss t=9.73 seconds, Particle. Will Return to its Initial position of (tos)

& for seteb : VE18 Suat 1&t&c Jidat =1&+ + c Boundary Conditioni- t=5s. x = 4,5m at 1 4.5 = 1815)+c > c= -85.5 - 18€ - 85.5 (x2+36) X - 3 22.5 m at t=6s. x = 1816) - 85.5 for 6. <+18 V = 36-0,5+ 2= Svar = √(32-0.5+²ydt 3 N = 36t tc V x= 3 LL.5 sa=7 Boundary condition in at 36(6) 22.5 05 3 16}+c c = -157.5 0.5 ² - 157.5 (52+28) x = 36€ (4 at t- & Second a = 3618) - 5 (839-15715 a u tlist

the from eq Plothing graph zimo 45.17 22.5m 4,5m 45 -S 75 8s 4.73 Second Gm -13.5 (S) 2 -2.25 +² o at se 4.5+2-274 +27 estas a = 184-85.5 9ラョラー 8 7759 34t-54-157.5 for Acceleration in dv dE (-4.54) = -4,5 mlje 0 < t <2 s3737 à 9 myse a (94-27) = D 9 377s > ale (18) dv at zolue 6<t< =) at the (36–05+) t mise

& Accelesation plot ! هر (۱ ) » 64 با لاي ) + AS کد اوکے یک لو 33 کی 4- 25 عي( - - - - . ومنها - -و بلا لام

Hope you got the concept well...!!!!