Yefim S. answered 05/18/21
Math Tutor with Experience
(a) s= ∫03(5t - 9)dt = (5t2/2- 9t)03 = 45/2 - 27 = - 4.5 m
(b) Distance d = ∫03I5t - 9Idt = ∫09/5(9 - 5t)dt + ∫9/53(5t - 9)dt = (9t - 5t2/2)09/5 + (5t2/2 - 9t)9/53 = 81/5 - 81/10 +
45/2 - 27 - 81/10 + 81/5 = 81/5 - 9/2 = 16.2 - 4.5 = 11.7 m
Mary A.
Thank you so much!05/18/21