An object's two dimensional velocity is given by #v(t) = ( sin(pi/3t) , 2cos(pi/2t )- t )#. What is the object's rate and direction of acceleration at #t=5 #?

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Answer 1 · 2018-01-04T10:04:27

The rate of acceleration is #=4.17ms^-2# in the direction #97.2^@# anticlockwise from the #"x-axis"#

The acceleration is the derivative of the velocity

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

The acceleration is

#a(t)=v'(t)=(pi/3cos(pi/3t), -pisin(pi/2t)-1)#

When #t=5#

#a(5)=v'(5)=(pi/3cos(pi/3*5), -pisin(pi/2*5)-1)#

#=(0.52, -4.14)#

The rate of acceleration is

#||a(5)||=sqrt((0.52)^2+(-4.14)^2)=4.17ms^-2#

And the direction is

#theta=arctan(-4.14/0.52)=97.2^@# anticlockwise from the #"x-axis"#