The equation of motion of a particle is s = t3 − 3t, where s is in meters and t is in seconds. (Assume t ≥ 0.) (a) Find the velocity and acceleration as functions of t. v(t) = a(t) = (b) Find the acceleration after 3 s. m/s2 (c) Find the acceleration when the velocity is 0. m/s2

a) , b) , c)

Step-by-step explanation:

a) The velocity and acceleration function are obtained by deriving the displacement function once and twice, respectively:

b) The acceleration of the particle at such instant is:

c) The time when velocity is zero is:

The acceleration is:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.