Finding Velocity of Particle: Problem Solved

[tennistudof09]

I wanted to make sure I did this problem correctly. The problem is:

A particle moves along a straight line and its position at time t is given by s(t)=2t^3 - 21t^2 + 60t where s is measured in feet and t in seconds.

Find the velocity of the particle when t=0.

I took the derivative of s(t) and got 6t^2 - 42t + 60, and then substituted 0 in for t. I got 60ft/sec for the answer. Is this correct??

 

[bblenyesi]

\frac{ds}{dt}=v(t)

\frac{dv}{dt}=a(t)

Where s(t) is the displacement vector, v(t) is the velocity vector and
a(t) is the acceleration vector.

If you look at it in any given point, so I'm guessing that it's correct.