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 #?

Answer 1

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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Answer 2

Amelia Adams

To find the object's acceleration at t=5, calculate the second derivative of the position function with respect to time. The rate of acceleration at t=5 is given by the second derivative, and the direction is determined by the corresponding unit vector.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.