An object's velocity is given by #v(t) = (2t^2 +t +1 , sin2t )#. What is the object's rate and direction of acceleration at #t=6 #?

1 Answer
Jul 16, 2017

Answer:

#a = 27.1# #"LT"^-2#

#theta = 3.58^"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) = 2t^2 + t + 1#

#v_y(t) = sin(2t)#

Finding the derivatives:

#a_x(t) = d/(dx) [2t^2 + t + 1] = 4t + 1#

#a_y(t) = d/(dx)[sin(2t)] = 2cos(2t)# (I'll assume this is in radians)

Plugging in #t = 6# (no units), we have

#a_x = 4(6) + 1 = 27# #"LT"^-2#

#a_y = 2cos(2(6)) = 1.69# #"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(27^2 + 1.69^2) = color(red)(27.1# #color(red)("LT"^-2#

And the direction is

#theta = arctan((a_y)/(a_x)) = arctan(1.69/27) = color(blue)(3.58^"o"#

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