An object's two dimensional velocity is given by v(t) = ( t-2t^3 , 4t^2-t). What is the object's rate and direction of acceleration at t=3 ?
1 Answer
The rate of acceleration (unknown units) is
The direction is at a bearing
Explanation:
The velocity function is:
bbv(t) = ( t-2t^3 , 4t^2-t)
Differentiating wrt
bba(t) = ( 1-6t^2 , 8t-1)
When
bba(t) = ( 1-6*9 , 8*3-1) = (-53,23)
We can represent this by a vector:
bba(3)= -53bbhati + 23bbhatj 
So we gain the rate using Pythagoras:
|bba(3) |= |-53bbhati + 23bbhatj |
\ \ \ \ \ \ \ \ \ = sqrt(-53^2+23^2)
\ \ \ \ \ \ \ \ \ = sqrt(2809+529)
\ \ \ \ \ \ \ \ \ = sqrt(3338)
\ \ \ \ \ \ \ \ \ = 57.78 (2 \ dp)
And for the direction using Trigonometry:
tan alpha = 23/53 => alpha = 23^o (nearest degree)
So the direction is at a bearing
