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).