An object's two dimensional velocity is given by #v(t) = ( t-2 , 2cos(pi/2t )- 3t )#. What is the object's rate and direction of acceleration at #t=1 #?

1 Answer
Jun 26, 2017

Answer:

#a = 6.22# #"m/s"^2#

#theta = -80.8^"o"#

Explanation:

We're asked to find the magnitude (rate) and direction of the object's acceleration at #t = 1# #"s"#, given the object's velocity equation.

We need to find the object's acceleration as a function of time, doing so via differentiation of the velocity equation:

#veca = d/(dt) vecv#

#vecv = (t-2)hati + (2cos(pi/2t)-3t)hatj# (given, component form)

#v_x = t-2#

#a_x = 1# (derivative of #t-2#)

#v_y = 2cos(pi/2t)-3t#

#a_y = -pisin(pi/2t)-3# (derivative)

Now that we have the components of the acceleration as a function of #t#, let's plug in #1# for #t# to find the acceleration components at #t = 1# #"s"#:

#a_x = color(red)(1# #color(red)("m/s"^2#

#a_y = -pisin(pi/2(1))-3 = color(green)(-6.14# #color(green)("m/s"^2#

The magnitude of the acceleration at this time is

#a = sqrt((a_x)^2 + (a_y)^2) = sqrt((color(red)(1)color(white)(l)color(red)("m/s"^2))^2 + (color(green)(-6.14)color(white)(l)color(green)("m/s"^2))^2)#

#= color(blue)(6.22# #color(blue)("m/s"^2#

The direction of the acceleration is

#theta = arctan((a_y)/(a_x)) = arctan((color(green)(-6.14)color(white)(l)color(green)("m/s"^2))/(color(red)(1)color(white)(l)color(red)("m/s"^2))) = -1.41"rad" = color(purple)(-80.8^"o"#