Kinematics - find velocity and acceleration

[UrbanXrisis]

A particle moves along the x-axis accoring to the equation x=2.00+3.00t-t^2 where x is in meters and t is in time. At 3.00s what is the velocity and acceleration. For velocity, I would just take the derivative of x=2.00+3.00t-t^2 and sub in 3.00s and for the acceleration I would have to get the second derivative of x=2.00+3.00t-t^2 and sub in 3.00 seconds. Is this correct?

 

[Sirus]

I believe so. Hurray for calculus!:)

 

[M98Ranger]

Yes, your approach is correct. To find the velocity at 3.00 seconds, you would take the derivative of x=2.00+3.00t-t^2 and substitute t=3.00 seconds. This would give you the instantaneous velocity at that specific time.

Similarly, to find the acceleration at 3.00 seconds, you would take the second derivative of x=2.00+3.00t-t^2 and substitute t=3.00 seconds. This would give you the instantaneous acceleration at that specific time.

It is important to note that in kinematics, the derivative of position with respect to time gives velocity, and the second derivative gives acceleration. So your approach is correct in finding the velocity and acceleration at a specific time.