one last calculus problem (acceleration, from position fcn)

I

integragirl

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suppose the position (in meters) of a particle after t seconds is given by s(t)=t^3-10.5t^2-2t

a. find the equation for the velocity, v at time t
b. when does (at what t value) the particle reach a velocity of 52 m/s
c. find the equation for the acceleration, a at time t
d. what is a(1)


Any help would be greatly appreciated
 
Intea, the first derivative of the position equation is an equation for velocity and the second derivative is an equation for acceleration.

Should be a breeze for you to find the answers to your questions using that :p

Oh yeah just to help you see where that comes from, velocity is change in displacement divided by change in time (ds/dt) and acceleration is change in velocity over change in time (dv/dt). We know that the derivative gives us the slope which is the rate of change therefore the derivative of the position function gives us change in position with respect to change in time which is what velocity is defined as being. Same line of reasoning for acceleration.
 
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