Question

1. At time t, the position of a body moving along the s-axis is s= t³ – 9t² +24t m.

Give formulas for the body’s velocity and acceleration at time t.

Find the body’s acceleration each time the velocity is zero.

Find the body’s speed each time the acceleration is zero.

Answer 1

1. we of and acceleration know ef s(t) represents the position a particle in time t, Then 0 410))v(t) represents the velocity of that particle is time to 6 d fult)) = a(4) represents the of that particle in time to have, s(t) = t - 92² +242 diff. both sides with respect to t, we get sct)} = (22 94*+291) : vt) Lt) - 90 (2) + 24.07€) NOW, at 2 = 37 - 9x2t + 24x1 - 3t²18t+ 24 Therefore, The velocity of the B body is lu(t) = 3t² 18t+29 How, diff. ult) with respect to t, we get 10} } 3+ 18+ + 24 30() 184 ) + (24) It alt) It
The answer sheet has three pages.it is the first page
..at)= 3x2t - 18x1 + 0 6t- 18. body is Therefore, the acceleration of the la(t) = Gt-18 zero means - Now, The velocity becomes VED=0 32² 18t+24=0 or 3/t 6t+8)=0 +²67 +8=0 -4) (6-2)=0 t= 2, t = 4. "The velouty of the bochy will be zeso t=2 and t= a. Oy at t=2 Now at t=2, Then alt) = 6x2-18 = 12-18 = -6 and at t= 4, Then cast) = 6x4-18 = 24-18 = 6
second page
6 is moving Therefore, The body's acceleration al- time t = 2 is meters/ see? Here negative sign indicates the body negative acceleration iie in retardation, And the body's acceleration at time t=4 meters/ see? en es 6 The acceleration of the body becomes zero means act) = 0 Thun 6t-18=0 6t=18 I t= 3. The speed of the body is given by (t) = 1 3+2 18t+24) Now, when t = 3, Then |va) ) = 1/3 x 832 - 18x3 + 24 = /27 - 49 + 24) |-3) = 3 Therefore, the body's speed at t = 3 is meters / sell.
Third /last page