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=5 #?

1 Answer
A08
Mar 13, 2017

Given that object's two dimensional velocity is
#v(t)=(tsin(π/3t),2cos(π/2t)−3t)#.

Acceleration #a(t)=dotv(t)=d/dt(tsin(π/3t),2cos(π/2t)−3t)#
#=>a(t)=(tpi/3cos(π/3t)+sin(pi/3t),-2pi/2sin(π/2t)−3)#

Required acceleration at #t=5# is
#a(5)=((5pi)/3cos((5π)/3 )+sin((5pi)/3 ),-pisin((5π)/2 )−3)#
#=>a(5)=((5pi)/3xx1/2-sqrt3/2,-(pi+3))#
#=>a(5)=(1.75197, -6.14159)#

#|a(5)|=sqrt((1.75197)^2+(-6.14159)^2)#
#|a(5)|=6.3866" units"#

Direction is given by
#alpha=tan^-1(-6.14159/1.75197)=-74.08^@#

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