An object's two dimensional velocity is given by #v(t) = ( t^2 - 2t , cospit - t )#. What is the object's rate and direction of acceleration at #t=5 #?

1 Answer
Jul 16, 2017

Answer:

#a = 8.06# #"LT"^-2#

#theta = -7.13^"o"#

Explanation:

We're asked to find the the magnitude (rate) and direction of an object's acceleration at a given time, given its velocity equation.

To do this, we need too differentiate the velocity component equations, which are given as

#v_x(t) = t^2 -2t#

#v_y(t) = cos(pit) - t#

Finding the derivatives:

#a_x(t) = d/(dx) [t^2 -2t] = 2t-2#

#a_y(t) = d/(dx)[cos(pit)-t] = -pisin(pit) - 1# (I'll assume this is in radians)

Plugging in #t = 5# (no units), we have

#a_x = 2(5)-2= 8# #"LT"^-2#

#a_y = -pisin(pi(5))-1 = -1# #"LT"^-2#

(The #"LT"^-2# term is the dimensional form of the units for acceleration (#"distance"xx"time"^-2#). I used this term here since no units were given.)

The magnitude of the acceleration is thus

#a = sqrt((a_x)^2 + (a_y)^2) = sqrt(8^2 + (-1)^2) = color(red)(8.06# #color(red)("LT"^-2#

And the direction is

#theta = arctan((a_y)/(a_x)) = arctan((-1)/8) = color(blue)(-7.13^"o"#

Always make sure your arctangent calculation is in the right direction; it could be #180^"o"# off!