Question

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.

Answer 1

Answer:

$v=0.04m/s$

$s=-28.592m$

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:

$s=-28.592m$