The distance of a particle from its starting point at time t

The distance of a particle from it\'s starting point at time t second is given by 8 = 80t - 16t^2

what is the velocity of the particle after 2 seconds?

Solution

Velocity is rate of change of displacement:

Given : Displacement

D = - 16t² + 80t - 8
  
Velocity is the derivative of displacement:

V = dD/ dt = -32t + 80
  
V(2) = -32(2) +80 = 16