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
Aug 16, 2017

Answer:

#a = 4.12# #"LT"^-2#

#theta = 14.0^"o"#

Explanation:

We're asked to find the magnitude and direction of the acceleration of an object at #t = 1#, given the velocity component equations.

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 #t = 1#:

#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).