An object's velocity is given by #v(t) = (2t^2 -2t +1 , sint )#. What is the object's rate and direction of acceleration at #t=3 #?

1 Answer
Jul 17, 2017

Answer:

#a = 10.0# #"LT"^-2#

#theta = -5.65^"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 - 2t + 1#

#v_y(t) = sin(t)#

Finding the derivatives:

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

#a_y(t) = d/(dx)[sint] = cost# (I'll assume this is in radians)

Plugging in #t = 3# (no units), we have

#a_x = 4(3) - 2 = 10# #"LT"^-2#

#a_y = cos(3) = -0.990# #"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(10^2 + (-0.990)^2) = color(red)(10.0# #color(red)("LT"^-2#

And the direction is

#theta = arctan((a_y)/(a_x)) = arctan((-0.990)/10) = color(blue)(-5.65^"o"#

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