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

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Answer 1 · 2018-01-09T08:37:39

The rate of acceleration is #=sqrt(5)ms^-2# in the direction #=153.4^@# anticlockwise from the #"x-axis"#

The acceleration is the derivative of the velocity

#v(t)=(1/t, t^2)#

Therefore,

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

When #t=1#, the acceleration is

#a(1)=(-1,2)#

The rate of acceleration is

#||a(1)||=sqrt((-1)^2+(2)^2)=sqrt(5)#

The direction of acceleration is

#theta=arctan(-1/2)=153.4^@# anticlockwise from the #"x-axis"#