Question

Velocity of a particle is expressed as
Vx=39-8t²m/s
(a)=find average acceleration
(b)=instantaneous acceleration

Answer 1

(a) The average acceleration

$$a_x=\frac{v_2-v_1}{t_2-t_1}=\frac{(39-8t_2^2)-(39-8t_1^2)}{t_2-t_1}=-8(t_2+t_1).$$

(b) The instantaneous acceleration

$$a_x(t)=v_x'(t)=(39-8t^2)'=-16t.$$