A particle is moving in a vertical line whose equation of motion is given by `s = 32 + 46t+9t^2` where 's' is measured in meters and 't' is measured in seconds. Find the velocity and acceleration of the particle at t=3 sec.

Explanation

Let (T, D) be any point on the, where D denotes the distance at time T. Therefore, points (0, 2), (3, 8) and (T, D) are collinear so that
`(8-2)/(3-0) = (D-8)/(T-3)` or `6(T-3)=3(D-8)`
or, D = 2(T + 1)
which is the required relation.