Tangential Accel and Normal Accel.

1. The position of a particle in motion in the plane at time http://webwork2.math.utah.edu/webwor...5b59f2bc11.png is http://webwork2.math.utah.edu/webwor...56deb6f101.png
At time http://webwork2.math.utah.edu/webwor...388a5f95d1.png, determine the following:
(a) The speed of the particle
http://webwork2.math.utah.edu/webwor...d590dce2a1.png 8.85663593
(b) The unit tangent vector to http://webwork2.math.utah.edu/webwor...d0a4b001e1.png
http://webwork2.math.utah.edu/webwor...33a1f823c1.png .9936053 http://webwork2.math.utah.edu/webwor...04cdc92501.png .112909688 http://webwork2.math.utah.edu/webwor...237541f941.png

(c) The tangential acceleration
http://webwork2.math.utah.edu/webwor...4a7acc0aa1.png?
(d) The normal acceleration
http://webwork2.math.utah.edu/webwor...50c2302c71.png?
I figured out the speed and T(t) but i have no idea how to find the aT and aN. Please help out.

you're answer to b is correct but it is T(0) not T(t)

This is not going to be very clean anyway you approach it--so have a cup of coffee and a cigarette

if s is the speed = |v|


It might be easiest to compute aN = k(ds/dt)^2

then aT = sqrt( |a|^2 - aN^2)

Where k = |T'|/|r'|



Normally you can compute the components by :

if s is the speed = |v|

Then aT = d^2s/dt^2

Without having to compute the curvature you can compute

aN = sqrt( |a|^2 - aT^2)

where a = (8.8)^2 e^(8.8t) i + e^t j

However for this problem d^2s/dt^2 is very cumbersome