Question

P2. The motion of a partical moving in straight line is given by th
S=t^3-3t^2 +2t+5 where 's' is the displacement in metres and 't' is timein sec
Determine 1) velocity and acceleration after 4 sec. (1) maximum or minimum velocity
and corresponding displacement and (ii) time at which velocity is zero.​

Answer 1

Answer:

solved

Explanation:

S=t^3-3t^2 +2t+5

S(t) = t³ - 3t² + 2t + 5

v(t) = 3t² - 6t  + 2  , v(4) = 3x16 - 6x4 + 2 = 26m/s

a(t) = 6t - 6  , a(4) = 6x4 - 6 = 18m/s²

min. velocity at a(t) = 6t - 6 = 0 , t = 1 , v(1) = 3 - 6 + 2 = -1 m/s

corresponding displacement = S(1) = 1 - 3 + 2 + 5 = 5 m.

v(t) = 3t² - 6t  + 2 = 0 , t² - 2t  + 2/3 = 0

t = 2 ±√4/3 ÷2 = 1 ± 1/√3 sec.