An object's two dimensional velocity is given by #v(t) = ( t^3, t-t^2sin(pi/8)t)#. What is the object's rate and direction of acceleration at #t=12 #?

1 Answer
Jul 31, 2017

Answer:

#a = 462# #"LT"^-2#

#theta = -20.8^"o"#

Explanation:

We're asked to find the magnitude and direction of the acceleration of an object at a certain time, given the velocity equations as functions of time.

To do this, we need to find the acceleration of the object as a suction of #t#, by differentiating the velocity component equations.

We're given:

#v_x(t) = t^3#

#v_y(t) = t-t^2sin(pi/8)t#

So

#a_x(t) = d/(dt) [t^3] = ul(3t^2#

#a_y(t) = d/(dt) [t-t^2sin(pi/8)t] = ul(1-3t^2sin(pi/8)#

Now, we plug in #t = 12#:

#a_x = 3(12)^2 = ul(432#

#a_y = 1-3(12)^2sin(pi/8) = ul(-164#

The magnitude of the acceleration is thus

#a = sqrt((a_x)^2 + (a_y)^2) = sqrt((432)^2 + (-164)^2)#

#= color(red)(ul(462color(white)(l)"LT"^-2#

(the #"LT"^-2# term is the dimensional form of the units for acceleration; I used it here since no units were given)

The direction is

#theta = arctan((a_y)/(a_x)) = arctan((-164)/(432)) = color(blue)(ul(-20.8^"o"#

Always be sure to check your arctangent calculation, as it could be #180^"o"# off!