Finding normal and tangential acceleration

[Inveritatem]

Homework Statement

A point moves along the curve y = x³ + x such that the vertical component of velocity is always 3. Find the tangential and normal components of acceleration at the point P(2,10).

Homework Equations

Tangential Acceleration - a_T(t) = v(t) ⋅ a(t)/ magnitude of velocity vector
Normal Acceleration - a_N(t) = magnitude of (v(t) × a(t)) / magnitude of (v(t))

The Attempt at a Solution

The fact that they gave the VERTICAL component of constant velocity is really throwing me off. In class we have done some similar procedure but the HORIZONTAL component of velocity being given instead.

So if I get f(t) as the parametic equation for x and g(t) for the parametric equation for y, I thought I would have to get x in terms of y (ƒ(t) in terms of g(t)), so I would have to get the inverse of this function. But this requires something called Cavadro's Method and results in weird values I don't understand.

So I thought maybe I would keep the x and y values in terms of f(t). However, that results in some very weird x component values for the velocity and acceleration vectors for the cross product such as -2/9*cuberoot(3/t^5) - 3, so I feel like this is a mistake.

Anyone have any insights into this problem I'm not seeing?

 

[Inveritatem]

Can someone please move this to the physics homework forum? I think it wouldve been more appropriate for me to post over there.

 

[andrewkirk]

You know that $y=3t+C$ so the vertical component of acceleration is always zero, hence you only need to calculate the horizontal component.

Start with the equation $3t+C=x^3+x$. Differentiate both sides with respect to $t$, then re-arrange to get a formula for $\frac{dx}{dt}$ in terms of $x$. Then differentiate that wrt $t$ to get a formula for $\frac{d^2x}{dt^2}$ in terms of $x$. Then substitute $x=2$ into those two formulas to find the velocity and acceleration at the point (2,10).

 

[Inveritatem]

You know that $y=3t+C$ so the vertical component of acceleration is always zero, hence you only need to calculate the horizontal component.

Start with the equation $3t+C=x^3+x$. Differentiate both sides with respect to $t$, then re-arrange to get a formula for $\frac{dx}{dt}$ in terms of $x$. Then differentiate that wrt $t$ to get a formula for $\frac{d^2x}{dt^2}$ in terms of $x$. Then substitute $x=2$ into those two formulas to find the velocity and acceleration at the point (2,10).

Thank you so much! I didn't get what you were saying at first but your edit made it very clear. Thank you!