An object's two dimensional velocity is given by #v(t) = ( 5t-t^3 , 3t^2-2t)#. What is the object's rate and direction of acceleration at #t=4 #?

Question

1365 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-10T06:07:33

#(-43,22)#

Acceleration at #t = 4# is

#a_x(4) = d/dt(v_x) = 5 - 3t^2 = 5 - (3 × 4^2) = 5 - 48 = -43#

#a_y(4) = d/dt(v_y) = 6t - 2 = (6 × 4) - 2 = 22#

#a = (-43, 22)#

Direction:

#Tan θ = a_y/a_x = 22/(-43)#

#θ = Tan^-1(-22/43)#

Where #theta# is angle with #x#-axis