Motorcycle Acceleration: Find Position, Velocity & Max Velocity

[Musicman]

The acceleration of a motorcycle is given by ax(t) = At - Bt2, where A = 1.70 m/s3 and B = 0.170 m/s4. The motorcycle is at rest at the origin at time t = 0.

(a) Find its position as a function of time t.

Find its velocity as a function of time t.

b. Calculate the max velocity it attains. I already calculated the max velocity it attains which is 28.33 m/s.

 

[Hootenanny]

Hi there musicman and welcome to PF,

What are your thoughts on this question? Would you mind posting what you have attempted thus far?

 

[Mindscrape]

For the first question it looks like you are just supposed to integrate and sketch some graphs, but, as Hootenanny has already asked, what have you done so far?

 

[Musicman]

ok i got:

v - v' = integral(At - Bt^2)dt
v = (1/2)At^2 - (1/3)Bt^3 + v'

x - x' = integral( v dt )
x = (1/6)At^3 - (1/12)Bt^4 + v't + x'

 

[Hootenanny]

ok i got:

v - v' = integral(At - Bt^2)dt
v = (1/2)At^2 - (1/3)Bt^3 + v'

That's correct, as the motocycle starts from rest your v' or constant of integration drops out.

x - x' = integral( v dt )
x = (1/6)At^3 - (1/12)Bt^4 + v't + x'

This is also correct, again as your motocycle begins from rest at the origin your terms v' and x' drop out.

 

[Musicman]

so would i write those exact functions in the space provided? or solve for an answer? and do i leave the apostrophes after the variables or no

 

[Hootenanny]

so would i write those exact functions in the space provided? or solve for an answer? and do i leave the apostrophes after the variables or no

Well, they can be simplified a bit, but will your tutor accept unsimplified answers? As a said before, the v', v't and x' terms drop out because of the intial conditions imposed by the question;

The motorcycle is at rest at the origin at time t = 0

 

[Musicman]

yea, they got to be simplified, thanks for the help.

 

[Hootenanny]

yea, they got to be simplified, thanks for the help.

Have you attempted to simplify them? How are you getting on with your second question?

 

[Musicman]

v = (1/2)At^2 - (1/3)Bt^3 + v' was this the one as position as a function of time or velocity? the velocity right? because x is displacement

 

[Hootenanny]

v = (1/2)At^2 - (1/3)Bt^3 + v' was this the one as position as a function of time or velocity? the velocity right? because x is displacement

This is a function of velocity, this is a result of integrating acceleration.

 

[Musicman]

ok right after it says find its velocity as a funtion of time it says (m/s) after the blank, so are they askign for a number or just the equation v=(1/2)At^2-(1/3)Bt^3...and isn't it ok for me to take off the +v' sicne it is zero?

 

[Hootenanny]

ok right after it says find its velocity as a funtion of time it says (m/s) after the blank, so are they askign for a number or just the equation

Is a function a number of an equation?

v=(1/2)At^2-(1/3)Bt^3...and isn't it ok for me to take off the +v' sicne it is zero?

Yes, in fact you should do as I have said on numerous occasions previously.

 

[Musicman]

i wrote an actual number for position as a function of time and velocity as a function of time and got it wrong

 

[Hootenanny]

i wrote an actual number for position as a function of time and velocity as a function of time and got it wrong

Damn webassign homework . Sorry, I'm not getting at you. What exactly did you write?

 

[Musicman]

13.12 m for the position as a function of time and 17.34 m/s as the velocity as a function of time.

 

[Hootenanny]

13.12 m for the position as a function of time and 17.34 m/s as the velocity as a function of time.

Where did you get those numbers from? There is no way you can calculate them; besides the question asks for a function, not a numerical answer.