An object's two dimensional velocity is given by #v(t) = ( 2t^2 , -t^3 +4t )#. What is the object's rate and direction of acceleration at #t=7 #?

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Answer 1 · 2017-05-13T09:13:41

The rate of acceleration is #=145.72ms^-1# and the direction is #=169.3#º anticlockwise from the x-axis.

The acceleration is the derivative of the velocity.

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

#a(t)=v'(t)=(4t, -3t^2+4)#

Therefore,

#a(7)=(4*7, -3*49+4)=(28,-143)#

The rate of acceleration is

#||a(4)||=sqrt(28^2+(-143)^2)#

#=sqrt21233#

#=145.72ms^-2#

The direction is

#theta=arctan(-28/143)#

#theta# lies in the 2nd quadrant

#theta=180-10.7=169.3º#