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An object is moving with velocity (in ft/sec) v(t) : t2 + 4t 5. (a) Find the displacement and total distance travelled from t : 0 to t : 6 Displacement: d' (b) Total Distance Travelled: By the net change theorem, 6 (a) Total displacement = v(t) dt 6 = tz + 4t - 5) dt +2+1 tl+1 6 = 2+ 1 + (4 ) 5t 1+1 t=0 43 6 = + 2t2 - 5t Jt=0 63 3 - + (2) (62) - (5)(6) - [03 + (2) (02) -(5) (0)]= 114 (b) Total distance traveled = Given that at t = 0, v(0) = - 5. So the velocity is negative to start and the graph of y = v(t) will be below the t-axis. We need to find when the object changes direction. Solve v(t) = 0 t2 + 4t - 5 = 0 (t + 5)(t - 1) = 0 t = - 5,1 Since t > 0 the object changes direction when t = 1 Thus, Since v(t) 0 from t > 1 tot = 6 we get 1 10 (1) at1 = 10 - v(t) dt + v(t) dt And 1, u(t)at - (that - v(t)tat6 And v(t) dt - v(t)tat - J v(t)tat Thus, lo (t) at1 - - [ v(t)at + J (that - J v(t)tat 6 6 to(t) del - J v(t)tatt - 2 / v(t) at From above we get v(t)tdt = 114, so we have 6 lv(t) dt| = 114 - 2 v(t) dt 6 o lv(t) dt) = 114 - 2/ (t2 + 4t - 5) dt 6 lv(t) dt| = 114 - t2+ 1 tl+1 2+ 1 + ( 4) 1+1 - 5t It=0 6 13 lv(t)dt) = 114 - 2 o + 2t2 - 5t It =0 6 13 lv(t) dt = 114 - 2 + (2) (12) - (5)(1) + 2 03 + (2) (02) - (5) (0)]6 13 v(t) dt = 114 - 2 0 + (2) (12) - (5)(1) + 2[03 + (2) (02) - (5) (0) 6 lv(t) dt| = 114 - (2) w | 0o 0 + (2) (0) 6 16 lv(t) dt| = 114 + 0 3 6 358 lv(t) dt = ~119.333 3