Question

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.

18 15 12 9 6 velocity (ft/s) v = 36-0.52 (from t=6 to t=8) 0 -6 -9 a time (5)

Answer 1

a Реле, For oct 22 velocity varies linearly, V-oto -9 2 ds 4.5t 2 Hica (-4,54 - = - 4.5 o 4.5 (23² d=2 t=2 For 2 ef 55 » velocity vosies lineonly v=-9 to v= 18 ftls. V +9 t-2 ist g 5-2 vt t-2 3 27 vtg- It - 18 vagt - 27 & for 2 at 25 ds gt- 27 It 5 Gt - 27 dt كام 2 stas - svaz - (972 - 2787 2

(4.5 (5² – 27 (5) -[(4,5) 12² - 27 (2) (-9) = - 22.5+ 36 = 13.5 = 4.5 m ft 4.5 m ft For 52+26 velocity r= constant av = 18 . stoo-sing z vx t = 18x 1 sto 4.52 18 [shto = 225ft Fur 6 & LG va 036-0582 ds It = 36-0.5f? S36-0,58²2dt star-sto [36t- ost 3 22.5 = Staç - 225 (2668) - 0.593)-(366) -e5.com = 22,5 + 202, 67 - Iso stes thea = 45.17 ft

3. s- & graph in + For oct ca Subs - 4.5$ Z 45.13 45 До 35 & For 25t5 30 27.5 SU) = 4,51 - 27t-g 20 15 10 (At) 4.5 For 54fc6 b 1 2 3 4 5 7 - 10 9 sch)518 18£ +4,5 Plus - 20. & tcs) - 648 as sets = 360 - 0.5t² + 22.5 3 H Now, a-t graph in کد های for occa a- dv 2) a= -9 2 을 = - 4,5 It 2 ct 25 (a= 9 For 5st s6 а со (reconstant For 651 58 a=- t a = dv at 10 9 g 1 6 a 2 oft 3 4 7 05 8 -5 -4.5 -10 time (s)

4 Q1.2 Function of accelestion over 6ct as Ha=rt k Angoes 7 361 – 6, EP 72,5 Borse 3