How do I find the rate of change and acceleration in a vector problem?

[zwook]

My book says V=dx/dt=B+3Ct^2 when x=A+Bt+Ct^3. This doesn't make any sense to me? Could anybody explain how x/t turned into B+3Ct^2 ?

a=dV/dt=6Ct How can V/t turn into 6Ct ? I don't understand this either :( If anobody could explain this would help me really really much.

 

[CornMuffin]

My book says V=dx/dt=B+3Ct^2 when x=A+Bt+Ct^3. This doesn't make any sense to me? Could anybody explain how x/t turned into B+3Ct^2 ?

a=dV/dt=6Ct How can V/t turn into 6Ct ? I don't understand this either :( If anobody could explain this would help me really really much.

V is the rate of change of position, so you would have to take the derivative of x to get V, let's take the derivative of each component, dx/dt = (d/dt)A + (d/dt)Bt + (d/dt)Ct^3
A is a constant, therefore the derivative of A would become zero.
The derivative of anything in the form of Dt^n would equal Dnt^(n-1), so the derivative of Bt would equal B(1)t^(1-1), which simplifies to B
And the derivative of Ct^3 would equal C(3)t^(3-1), which simplifies to 3Ct^2
And when you put it all together you get, V = B + 3Ct^2

to find acceleration, you just take the derivative of velocity.
dv/dt = (d/dt)B + (d/dt)3Ct^2
The derivative of B, a constant, would be zero, and the derivative of 3Ct^2 would be 3(2)Ct^(2-1), which would simplify to 6Ct
So acceleration equals 6Ct


hope this helps

 

[zwook]

V is the rate of change of position, so you would have to take the derivative of x to get V, let's take the derivative of each component, dx/dt = (d/dt)A + (d/dt)Bt + (d/dt)Ct^3
A is a constant, therefore the derivative of A would become zero.
The derivative of anything in the form of Dt^n would equal Dnt^(n-1), so the derivative of Bt would equal B(1)t^(1-1), which simplifies to B
And the derivative of Ct^3 would equal C(3)t^(3-1), which simplifies to 3Ct^2
And when you put it all together you get, V = B + 3Ct^2

to find acceleration, you just take the derivative of velocity.
dv/dt = (d/dt)B + (d/dt)3Ct^2
The derivative of B, a constant, would be zero, and the derivative of 3Ct^2 would be 3(2)Ct^(2-1), which would simplify to 6Ct
So acceleration equals 6Ct




hope this helps

Thank you very very much!
Now I actually understand what I'm doing =)

Tommorow I'm going to study kinematics of spinning motion, hope I won't have trouble there!