Question

1. A body projected in a vertical y (altitude) is given by y = 3t²+ 2t + 5, where t is in seconds.
Calculate (i) the velocity at time t​

Answer 1

y = 3t²+ 2t + 5

Differentiating both sides w.r.t x

$\frac{dy}{dx} = 6t + 2$

Velocity at time t =  $\frac{dy}{dx} = 6t + 2$

Answer 2

y = 3t²+2t+5

dy/dt = 6t+2

bt differentiating both side w.r.t t