Finding Velocity and Acceleration for a Moving Particle

[tcc88]

A particle moves along the x-axis according to the equation
x = 1.95 + 2.97t − 1.00t^2, where x is in meters and t is in seconds.

(a) Find the position of the particle at t = 3.30 s.
(b) Find its velocity at t = 3.30 s.
(c) Find its acceleration at t = 3.30 s.

I must find the instantaneous acceleration and instantaneous velocity, I already found the position were the particle is at t = 3.30 s [0.861m] but I am having trouble finding the other two. I know I must get the derivative of the equation once for velocity and twice for acceleration to find the answer to this question but I am having trouble. What would the derivative if the equation for acceleration be? Wouldn't attempting to get the derivative twice for that equation yield just a number? I just stared Calc. and for some reason my Physics class is a little ahead of my Calc. I class in Calculus concepts -_-. So please help... I know this kind of question isn't hard, I just need a little bit of guidance.

Thank You.l

 

[cepheid]

Welcome to PF,

A particle moves along the x-axis according to the equation
x = 1.95 + 2.97t − 1.00t^2, where x is in meters and t is in seconds.

(a) Find the position of the particle at t = 3.30 s.
(b) Find its velocity at t = 3.30 s.
(c) Find its acceleration at t = 3.30 s.

I must find the instantaneous acceleration and instantaneous velocity, I already found the position were the particle is at t = 3.30 s [0.861m] but I am having trouble finding the other two. I know I must get the derivative of the equation once for velocity and twice for acceleration to find the answer to this question but I am having trouble. What would the derivative if the equation for acceleration be? Wouldn't attempting to get the derivative twice for that equation yield just a number? I just stared Calc. and for some reason my Physics class is a little ahead of my Calc. I class in Calculus concepts -_-. So please help... I know this kind of question isn't hard, I just need a little bit of guidance.

Thank You.l

Just differentiate the function x(t) with respect to time once to get velocity. Then differentiate that to get acceleration.

Yes, in this case, the acceleration will be just a constant. This should not be surprising, because the position function is quadratic (2nd order). You can see that it has exactly the same form as the constant acceleration distance vs. time equation from kinematics that you are probably already familiar with.

EDIT: to clarify: A linear (1st order) function has a constant slope (derivative)

A quadratic (second order) function has a derivative that is linear (first order), which means that ITS derivative (the second derivative) is a constant.

Every time you differentiate a polynomial, you reduce its order by 1.

 

[tcc88]

Welcome to PF,



Just differentiate the function x(t) with respect to time once to get velocity. Then differentiate that to get acceleration.

Yes, in this case, the acceleration will be just a constant. This should not be surprising, because the position function is quadratic (2nd order). You can see that it has exactly the same form as the constant acceleration distance vs. time equation from kinematics that you are probably already familiar with.

EDIT: to clarify: A linear (1st order) function has a constant slope (derivative)

A quadratic (second order) function has a derivative that is linear (first order), which means that ITS derivative (the second derivative) is a constant.

Every time you differentiate a polynomial, you reduce its order by 1.

OK then thank you, I knew it wasn't that hard!