An object's two dimensional velocity is given by #v(t) = ( 3t^2 - 2t , t )#. What is the object's rate and direction of acceleration at #t=1 #?
1 Answer
Explanation:
We're asked to find the magnitude and direction of the acceleration of an object at
Acceleration is the first derivative of velocity, so we differentiate the velocity component equations:
#a_x(t) = d/(dt) [3t^2 - 2t] = ul(6t - 2#
#a_y(t) = d/(dt) [t] = ul(1#
Substituting in
#a_x = 6(1) - 2 = 4#
#a_y = 1#
The magnitude of the acceleration is given by
#a = sqrt((a_x)^2 + (a_y)^2) = sqrt(4^2 + 1^2) = color(red)(ulbar(|stackrel(" ")(" "4.12color(white)(l)"LT"^-2" ")|)# The
#"LT"^-2# is the dimensional form of the units for acceleration. I used it here because no units were given.
The direction is given by
#theta = arctan((a_y)/(a_x)) = arctan(1/4) = color(red)(ulbar(|stackrel(" ")(" "14.0^"o"" ")|)# measured anticlockwise from the positive
#x# -axis (which is the normal measurement standard for planar angles).
