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

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Answer 1 · 2018-01-10T14:22:19

The rate of acceleration is #=3.01ms^-2# in the direction #=266.6^@# anticlockwise from the #"x-axis"#

The acceleration is the derivative of the velocity.

#v(t)=(tsin(pi/3t), 2cos(pi/2t)-3t)#

Therefore,

#a(t)=v'(t)=(sin(pi/3t)+pi/3tcos(pi/3t), -pisin(pi/2t)-3)#

When #t=2#

The acceleration is

#a(2)=v'(2)=(sin(pi/3*2)+pi/3*2*cos(pi/3*2), -pisin(pi/2*2)-3)#

#a(2)=(sqrt3/2-pi/3, -3)=(-0.18,-3)#

The rate of acceleration is

#=||a(2)||=sqrt((-0.18)^2+(-3)^2)=3.01ms^-2#

The direction is #theta=180+arctan (3/0.18)=266.6^@# anticlockwise from the #"x-axis"#