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The acceleration of a particle is given by a = 3t – 4, where a is in meters per second squared and t is in seconds. Determine the velocity and displacement for the time t = 3.6 sec. The initial displacement at t = 0 is s0 = – 8 m, and the initial velocity is v0 = – 5 m/sec.

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Answer:


v=0.04m/s\\


s=-28.592m\\

Step-by-step explanation:


a = 3t-4


v(t)=\int\limits^t_0 {a(t)} \, dt =3/2*t^(2)-4t+v_0\\

if t=3.6s and initial velocity, v0, is -5m/s


v=0.04m/s\\


s(t)=\int\limits^t_0 {v(t)} \, dt =1/2*t^(3)-2t^(2)+v_0*t+s_0\\

if t=3.6s and the initial displacement, s0, is -8m:


s=-28.592m\\

User Ewgenij Sokolovski
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