An object's two dimensional velocity is given by #v(t) = ( t-2t^3 , 4t^2-t)#. What is the object's rate and direction of acceleration at #t=3 #?

1 Answer
Jan 29, 2018

Answer:

The rate of acceleration (unknown units) is # 57.78 \ (2dp)#

The direction is at a bearing #203^o#

Explanation:

The velocity function is:

# bbv(t) = ( t-2t^3 , 4t^2-t)#

Differentiating wrt #t# we get the acceleration function:

# bba(t) = ( 1-6t^2 , 8t-1)#

When #t=3# we get

# bba(t) = ( 1-6*9 , 8*3-1) = (-53,23) #

We can represent this by a vector:

# bba(3)= -53bbhati + 23bbhatj #

https://www.wolframalpha.com/input/?i=-53i%2B23j

So we gain the rate using Pythagoras:

# |bba(3) |= |-53bbhati + 23bbhatj |#
# \ \ \ \ \ \ \ \ \ = sqrt(-53^2+23^2)#
# \ \ \ \ \ \ \ \ \ = sqrt(2809+529)#
# \ \ \ \ \ \ \ \ \ = sqrt(3338)#
# \ \ \ \ \ \ \ \ \ = 57.78 (2 \ dp)#

And for the direction using Trigonometry:

# tan alpha = 23/53 => alpha = 23^o# (nearest degree)

So the direction is at a bearing #203^o#