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=1 #?

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Answer 1 · 2017-12-18T06:26:41

The rate of acceleration is #=8.6ms^-2# in the direction #144.5^@# anticlockwise from the x-axis

The acceleration is the derivative of the velocity

#a(t)=(dv)/dt#

Here,

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

#a(t)=v'(t)=(1-6t^2,8t-1)#

Therefore,

When #t=1#

#a(1)=(1-6,8-1)=(-5,7)#

The rate of acceleration is

#||a(1)||=sqrt((-5)^2+(7)^2)=sqrt(25+49)=sqrt74=8.6ms^-2#

The direction of acceleration is

#theta=arctan(-5/7)=144.5^@# anticlockwise from the x-axis