Question

14.0 m/s [R] 5 mis?[R] bamy 5*[L] -10.0+ -20.0 + 1. The acceleration at t=2.999s Sn 2. The acceleration at t=20.5s 3. The displacement from t=0.0s to t=4.0s 4. The displacement from t=12.0s to t=32.0s 5. The average velocity from t=12.0s to t=32.0s 6. The object's velocity at 13.5 seconds. 7. The time interval when the object travelled in uniform motion 8. The time(s) when the object was at rest. 9. The time interval when the object is moving to the left and slowing down. 10. The time interval when the object is moving to the right and speeding up.

please answer the entire question (all the parts 1-10) for thumbs up.

thanks

Answer 1

thumbs up please

W utela i from graph :- oçtsu :- +80 = 90.+ Ś = st-40 :- y = fopen(tau) , y = - 7 Sad - 4-143 - t-12) 5 y = 14 – 1.47+ 1.2 = 15.2 - 1.4t ( 12 <t<29 Liu) da < t 32 - 4 = -80 (t-4) 44-dt 10 34<t< 36 , V = -20 at t= 2.999 5 5 from ac duz 5 la = 5 mise & at t=2015 S from ) - a-- du = -1.4 = la = -1.4 mise x = luidt .. - sist-20) dt from a = işt2 - 20t)” = - 40 - 80 -40m = x = 4 , dt + jvqdf = (IS.A - 1.4 ) dt + J44 – ) 4 - = 15.2*10 - 0.7x (227-12) + 44x10 - (322-222) - 152- 238 +410 - 540 - 216 m

Actually We are not allowed to attmpt more than 4 subparts of a perticular question