Particle Position and Time: Solving for Velocity and Acceleration

[Richard C.]

Homework Statement

he position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters and t in seconds.

(a) What dimension and units must c have?
m/s2
s2/m
m2/s
s/m2


What dimension and units must b have?
s/m3
m3/s
s3/m
m/s3


For the following, let the numerical values of c and b be 3.2 and 1.0 respectively.

(b) At what time does the particle reach its maximum positive x position?


(c) What distance does the particle cover in the first 4.0 s?


(d) What is its displacement from t = 0 to t = 4.0 s?


(e) What is its velocity at t = 1.0?

What is its velocity at t = 2.0?

What is its velocity at t = 3.0?

What is its velocity at t = 4.0 s?


(f) What is its acceleration at at t = 1.0 s?

What is its acceleration at at t = 2.0 s?

What is its acceleration at at t = 3.0 s?

What is its acceleration at at t = 4.0 s?

Homework Equations

I know that in the bold are correct. Unless of course I'm proven to be wrong on this of course.

So, that seems to use the grand dx/dt for (b) unless I'm incorrect. Though, I'm not sure entirely.

So, it would be as such unless I'm mistaken.

v = dx / dt = [(3)*35 + (2) 1]=0

Though, something tells me that I may be incorrect, if so can you please explain why in details. (Sorry to be demanding.) I feel it has to do with the coefficient.

 

[edziura]

For a) your 1st answer is incorrect, but the second is correct, assuming the numbers 2 and 3 are exponents.

For b), yes, set dx/dt = 0 and solve for t; I don't understand your answer.

x = ct² - bt³

\frac{dx}{dt} = 2ct - 3bt² = 0,

etc.

 

[Richard C.]

For a) your 1st answer is incorrect, but the second is correct, assuming the numbers 2 and 3 are exponents.

For b), yes, set dx/dt = 0 and solve for t; I don't understand your answer.

x = ct² - bt³

\frac{dx}{dt} = 2ct - 3bt² = 0,

etc.

I noticed that I made a mistake and its m/s^2. Thanks for that, though.

Ahh, I noticed I have to move the coefficient down and lessen it by one for the reminder. I presume you would then just fill in with the two numbers given. Am I doing this wrong so far?

Next you would get two derivatives, right?

So it would be a quadratic function, though am I missing anything?

 

[edziura]

I noticed that I made a mistake and its m/s^2. Thanks for that, though.

Ahh, I noticed I have to move the coefficient down and lessen it by one for the reminder. I presume you would then just fill in with the two numbers given. Am I doing this wrong so far?

That's correct.

Next you would get two derivatives, right?

I don't understand your question.

 

[Richard C.]

I don't understand your question.

Sorry.

Once you get: dx/dt=2(3.2)t-3(1)t^2=0 you would then find the derivative, right or not? Which would be a quadratic function. If so you would then solve that.

 

[edziura]

Sorry.

Once you get: dx/dt=2(3.2)t-3(1)t^2=0 you would then find the derivative, right or not? Which would be a quadratic function. If so you would then solve that.

dx/dt=2(3.2)t-3(1)t^2 is the derivative. Set it = to 0 and solve.