Question

Velocity of a particle is expressed as
Vx=16-9t²m/s
(a)=find average acceleration
(b)=instantaneous acceleration

Answer 1

$$16-9T^2=0\to T=\frac{4}{3}s$$

a)

$$a_{av}=0.5(-18)\frac{4^2}{3^2}=-16\frac{m}{s^2}$$

b)

$$a=\frac{dv}{dt}=0-9(2t)=(-18t)\frac{m}{s^2}$$