An object's two dimensional velocity is given by #v(t) = ( cost , -t^3 +4t )#. What is the object's rate and direction of acceleration at #t=7 #?

1 Answer
Jun 17, 2017

Answer:

#a = 143"m"/("s"^2)#

#theta = 270^"o"#

Explanation:

We're asked to find an object's magnitude and direction of the acceleration given its velocity equation.

We need to find the object's acceleration as a function of time, which we can do by differentiating the velocity equation.

The derivative of #t^n# is #nt^(n-1)#, and the derivative of #cos(at)# is #-asin(at)#, so let's use these to find the acceleration vs. time equation:

Velocity equation (component form):

#v(t) = (cost)hati + (-t^3 + 4t)hatj#

  • #d/(dt) (cost) = color(red)(-sint#

  • #d/(dt) (-t^3) = color(darkorange)(-3t^2#

  • #d/(dt) 4t = color(green)(4#

Therefore,

#a(t) = (color(red)(-sint))hati + (color(darkorange)(-3t^2) + color(green)(4))hatj#

Now, let's plug in #7# for #t# to find the acceleration at that time:

#a(7) = (-sin7)hati + (-3(7)^2 + 4)hatj#

#= -0.657hati - 143hatj#

The magnitude of the acceleration at this time is

#a = sqrt((a_x)^2 + (a_y)^2) = sqrt((-0.657)^2 + (-143)^2)#

#= color(blue)(143"m"/("s"^2)#

The direction of the acceleration is given by

#theta = arctan((a_y)/(a_x)) = arctan((-143)/(-0.657)) = 4.71"rad" = color(purple)(270^"o"#

Therefore, at #t = 7# #"s"#, the acceleration is #color(blue)(143"m"/("s"^2)#, at a direction of #color(purple)(270^"o"# anticlockwise from the positive #x#-axis.