Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. find the acceleration at 3 seconds. b. find the displacement at 2 seconds.

Explanation:

a )  V = 3 cos(0.5t)

differentiating with respect to t

dv /dt = -3 x .5 sin0.5t

= -1.5 sin0.5t.

acceleration = - 1.5 sin 0.5t

when t = 3 s

acceleration = - 1.5 sin 1.5

= - 1.496 ms⁻²

v = 3 cos.5t

b )  dx/dt = 3 cos 0.5 t

dx = 3 cos 0.5 t dt

integrating on both sides

x = 3 sin .5t / .5

x = 6 sin0.5t

At t = 2 s

x = 6 sin 1

x = 5.05 m

increases it

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