# Calc 3: Motion in xy plane question

1. Oct 1, 2009

### widmoybc

1. The problem statement, all variables and given/known data

In the xy plane, where distances are measured in feet, a 96 lb object is moving from left to right on the curve y=e^-x. At t-0 its speed is 6ft/s and is speeding up at (square root of 8) ft/s^2. what is the total force on it at that point? Give floating point values for the magnitude and angle of inclination of this force.

2. Relevant equations

F=ma

3. The attempt at a solution

I just don't really know where to begin. i'm sure once i get started on the right track, i can finish it. we know that the point in question is (0,1), and we know v(0)=6 ft/s and a(0)=sqrt8 ft/s^s. and we know y=e^-x, dy/dx= -e^-x, and second derivative= e^-x. the unit tangent is i-j and the unit normal is i+j. Now i'm stuck. do i need to use the breakdown of a into componets equation with the curvature and whatnot?

2. Oct 1, 2009

### HallsofIvy

Staff Emeritus
F= ma is a very good place to start! You know the object is moving on the curve $y= e^{-x}$ so its velocity vector must be tangent to that curve. And you have already calculated that the derivative is $y'= -e^{-x}$ so you know a vector tangent to that is $x\vec{i}- e^{-x}\vec{j}$ for each x. Further you are told that the speed (length of the velocity vector) is 6 ft/sec at t= t_0 and increases at the constant rate of $2\sqrt{2}$: the speed at time t is $6+2\sqrt{2}(t- t_0)$. That means that the velocity vector at time t is parallel to $x\vec{i}- e^{-x}\vec{j}$ (and so is of the form $xv\vec{i}- ve^{-x}\vec{j}$ which has length $v\sqrt{x^2- e^{-2x}})= 2\sqrt{2}$).

3. Oct 1, 2009

### widmoybc

Alright, thanks! i think i got my answer. i ended up getting an acceleration vector of 11sqrt2 i + 7sqrt2 j. now i just need to multiply that by the mass to get the vector of the force. do i need to convert pounds to slugs? then my total force will be in pounds?