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An object's two dimensional velocity is given by #v(t) = ( 3t^2 - 2t , 1- 3t )#. What is the object's rate and direction of acceleration at #t=4 #?

Answer 1

Julia Aguilar

The rate of acceleration is #=22.2ms^-1# in the direction of #=352.2º#

The acceleration is the derivative of the velocity.

#v(t)=(3t^2-2t, 1-3t)#

#a(t)=v'(t)=(6t-2, -3)#

Therefore,

#a(4)=(24-2, -3)=(22,-3)#

The rate of acceleration is

#||a(4)||=sqrt(22^2+(-3)^2)#

#=sqrt493#

#=22.2ms^-2#

The direction is

#theta=arctan(-3/22)#

#theta# lies in the 4th quadrant

#theta=360-7.8=352.2º#

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Answer 2

Julia Aguilar

To find the acceleration, take the derivative of the velocity function with respect to time, then evaluate it at t=4. The acceleration vector is given by a(t) = (6t - 2, -3). So, at t=4, the acceleration vector is a(4) = (22, -3). The rate of acceleration is the magnitude of the acceleration vector, which is √(22^2 + (-3)^2) ≈ 22.44 units per second squared. The direction of acceleration is the direction of the acceleration vector, which is approximately 272.4 degrees counterclockwise from the positive x-axis.

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Answer from HIX Tutor

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