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

1 Answer
Jul 6, 2017

Answer:

#a = 6.52# #"m/s"^2#

#phi = 259^"o"#

Explanation:

We're asked to find the object's acceleration at time #t = 1# #"s"# from a given velocity equation.

To do this, we need to find the object's acceleration components as a function of #t#, via differentiation of the velocity equations:

Finding the derivative of velocity component equations, we have

#v_x = e^t - 2t^2#

#v_y = t - te^2#

#a_x = d/(dt) [e^t - 2t^2] = e^t - 4t#

#a_y = d/(dt) [t - te^2] = 1 - e^2#

At time #t = 1# #"s"#, the acceleration components are thus

#a_x = e^((1)) - 4(1) = -1.28# #"m/s"^2#

#a_y = 1 - e^2 = -6.39# #"m/s"^2#

The magnitude (rate) of the acceleration is

#a = sqrt((-1.28color(white)(l)"m/s"^2)^2 + (-6.39color(white)(l)"m/s"^2)^2) = color(red)(6.52# #color(red)("m/s"^2#

The direction of the acceleration is

#phi = arctan((-6.39color(white)(l)"m/s"^2)/(-1.28color(white)(l)"m/s"^2)) = 78.7^"o"# #+ 180^"o" = color(blue)(259^"o"#

measured anticlockwise from the positive #x#-axis.

The #180^"o"# was added to fix the calculator angle error.