Expression of Acceleration and velocity

[ibysaiyan]

Homework Statement

The position of an object of mass m is given by:
http://imageshack.us/photo/my-images/62/20110920161443.jpg/
Find expression for the velocity and the acceleration and deduce the force components acting on the particle.

Homework Equations

W = vr, v = u+at , f = ma
v = d/t

The Attempt at a Solution

I was thinking of finding out the magnitude of the position (by squaring them followed by square root),so then I could divide position by time to get velocity then use SUVAT equation but so far I have achieved nothing sensible.. I would like to know am I on the right track ?

 

[Mark44]

Homework Statement

The position of an object of mass m is given by:
http://imageshack.us/photo/my-images/62/20110920161443.jpg/
Find expression for the velocity and the acceleration and deduce the force components acting on the particle.

Homework Equations

W = vr, v = u+at , f = ma
v = d/t

The Attempt at a Solution

I was thinking of finding out the magnitude of the position (by squaring them followed by square root),so then I could divide position by time to get velocity then use SUVAT equation but so far I have achieved nothing sensible.. I would like to know am I on the right track ?

I don't think so.

You have r as a (vector) function of t. v is defined as dr/dt, and a is defined as dv/dt.

 

[lanedance]

you can just differentiate the components to find the velocity (once) and acceleration (twice) components

I see no need for taking the magnitude, unless it asks for one

deduce the force using f=ma in component form

 

[ibysaiyan]

Thanks for your replies everyone!
Hm.. if I have understood the above posts correct then is this working of mine , right ?
r = (at+bt^2) i +4j + (Ce^-4t sin wt) k

Right, so by differentiating each components separately once i.e the first derivative should give me vector function of velocity.

component z becomes (at+bt^2) becomes: (t +2t) , 4J becomes J , and for the last component I use chain rule (un sure , I don't think product rule is appropriate)
(ce^-4t sinwt) k = (-4te^-4t cos w ) k Am I right? I am slightly confused about the 'constants' ...
hmm

 

[Mark44]

Thanks for your replies everyone!
Hm.. if I have understood the above posts correct then is this working of mine , right ?
r = (at+bt^2) i +4j + (Ce^-4t sin wt) k

Right, so by differentiating each components separately once i.e the first derivative should give me vector function of velocity.

component z becomes (at+bt^2) becomes: (t +2t) , 4J becomes J , and for the last component I use chain rule (un sure , I don't think product rule is appropriate)
(ce^-4t sinwt) k = (-4te^-4t cos w ) k


Am I right? I am slightly confused about the 'constants' ...
hmm

Most of what you wrote is incorrect.
For the x (not z) component, d/dt(at + bt²) \neq t + 2t.
For the y component, d/dt(4) \neq 1.
For the z component, you need to find d/dt(Ce^-4tsin(wt)). To do that, you will need to use the product rule, and then the chain rule (twice).

 

[ibysaiyan]

Most of what you wrote is incorrect.
For the x (not z) component, d/dt(at + bt²) \neq t + 2t.
For the y component, d/dt(4) \neq 1.
For the z component, you need to find d/dt(Ce^-4tsin(wt)). To do that, you will need to use the product rule, and then the chain rule (twice).

Thanks for the reply, could you tell me what exactly happens to the constants ?

 

[Mark44]

d/dt(K) = 0 and d/dt(a*f(t)) = a * f'(t)

 

[ibysaiyan]

d/dt(K) = 0 and d/dt(a*f(t)) = a * f'(t)

Sorry but I am further lost with the above expression.
Oh as for the derivative of 4j goes.. wouldn't that be 4 and a+2bt for i , so do the constants stay as they are ?
I think that makes sense since a 'constant' is any integer... for example if a was replaced by 3t then the derivative of that in respect to time would have been 3.

 

[Mark44]

d/dt(K) = 0 and d/dt(a*f(t)) = a * f'(t)

The first one says that the derivative of a constant is zero.

The second one says that the derivative of a constant times a function is the constant times the derivative of the function.

If you are finding derivatives of vector-valued functions, and you don't know these differentiation rules, you are in big trouble!

Sorry but I am further lost with the above expression.
Oh as for the derivative of 4j goes.. wouldn't that be 4 and a+2bt for i , so do the constants stay as they are ?

Look in your book at some examples of differentiating vector-valued functions. You don't actually differentiate the unit vectors i, j, and k. d/dt(4) \neq 4. For the other one, yes, d/dt(at + bt²) = a + 2bt

 

[Mark44]

I think that makes sense since a 'constant' is any integer

No, a constant is a fixed number. It could be an integer, but it doesn't have to be. For example, in the expression \pi x, \pi is a constant, but it isn't an integer.

... for example if a was replaced by 3t then the derivative of that in respect to time would have been 3.

But a is a constant, so it doesn't depend on any variable. That's why it's constant.

 

[ibysaiyan]

The first one says that the derivative of a constant is zero.

The second one says that the derivative of a constant times a function is the constant times the derivative of the function.

If you are finding derivatives of vector-valued functions, and you don't know these differentiation rules, you are in big trouble!
Look in your book at some examples of differentiating vector-valued functions. You don't actually differentiate the unit vectors i, j, and k. d/dt(4) \neq 4. For the other one, yes, d/dt(at + bt²) = a + 2bt

Thanks that makes much sense now.. sorry , I thought f and a above were force and acceleration.

No, a constant is a fixed number. It could be an integer, but it doesn't have to be. For example, in the expression \pi x, \pi is a constant, but it isn't an integer.
But a is a constant, so it doesn't depend on any variable. That's why it's constant.

Oh I see.. yea how stupid can I be... ijk denote Axis... That was very sloppy of me...