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

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Answer 1 · 2017-04-27T06:47:48

The rate of acceleration is #=4ms^-2# in the direction of #=73.6º#

Acceleration is the derivative of the velocity

#a(t)=v'(t)#

#v(t)=(3t-sqrtt,t^3-3t)#

#a(t)=v'(t)=(3-1/(2sqrtt),3t^2-3)#

When #t=2#,

#a(2)=(2.65,9)#

The rate of acceleration is

#||a(2)||=sqrt((2.65)^2+9^2)#

#=sqrt16=4#

The direction is

in the first quadrant

#theta=arctan(9/2.65)=73.6º#