Question

If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?

Answer 1

Answer:

v = -6m/s

Explanation:

$x=4-12t+3t^2$

$\frac{dx}{dt}=-12+6t$

For t = 1:

$\frac{dx}{dt}=-6$