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

Answer 1

#color(red)(a_x(2)=-0,184)#
#color(green)(a_y(2)=-1)#
#a(2)=1,017 #

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

#a_x(t)=d/(d t)(t sin(pi/3t))=1*sin(pi/3t)+t*pi/3cos(pi/3t) #

#a_x(2)=sin(2pi/3)+2*pi/3*cos(2pi/3)#

#a_x(2)=0,866+2*pi/3*(-1/2)#

#a_x(2)=0,866-pi/3#

#a_x(2)=0,866-1,05#

#color(red)(a_x(2)=-0,184)#

#a_y(t)=d/(d t)(2 cos(pi/2t)-t)#

#a_y(t)=-2*pi/2*sin(pi/2t)-1#

#a_y(t)=-pi*sin(pi/2t)-1#

#a_y(2)=-pi*sin(pi/2*2)-1#

#a_y(2)=-pi*sinpi-1" "sin pi=0#

#a_y(2)=-pi*0-1#

#color(green)(a_y(2)=-1)#

#a(2)=sqrt((a_x(2))^2+((a_y(2))^2))#

#a(2)=sqrt((-0,184)^2+(-1)^2))#

#a(2)=sqrt(0,034+1)#

#a(2)=sqrt(1,034)#

#a(2)=1,017 #

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Answer 2

To find the object's rate of acceleration at ( t = 2 ), we first need to find its acceleration vector, given by the second derivative of the position vector with respect to time.

The acceleration vector ( a(t) ) is the derivative of the velocity vector ( v(t) ) with respect to time:

[ a(t) = \frac{{dv(t)}}{{dt}} ]

Then, we evaluate ( a(t) ) at ( t = 2 ) to find the acceleration at that specific time.

Using the given velocity vector ( v(t) = (tsin(\frac{\pi}{3}t), 2cos(\frac{\pi}{2}t) - t) ), we find the acceleration vector:

[ a(t) = \left( \frac{d}{dt}(tsin(\frac{\pi}{3}t)), \frac{d}{dt}(2cos(\frac{\pi}{2}t) - t) \right) ]

Taking the derivatives, we get:

[ a(t) = \left( sin(\frac{\pi}{3}t) + \frac{\pi}{3}tcos(\frac{\pi}{3}t), -2sin(\frac{\pi}{2}t) - 1 \right) ]

Now, we evaluate ( a(t) ) at ( t = 2 ) to find the acceleration at ( t = 2 ):

[ a(2) = \left( sin(\frac{\pi}{3}(2)) + \frac{\pi}{3}(2)cos(\frac{\pi}{3}(2)), -2sin(\frac{\pi}{2}(2)) - 1 \right) ]

[ a(2) = \left( sin(\frac{\pi}{3}(2)) + \frac{\pi}{3}(2)cos(\frac{\pi}{3}(2)), -2sin(\pi) - 1 \right) ]

[ a(2) = \left( sin(\frac{2\pi}{3}) + \frac{2\pi}{3}cos(\frac{2\pi}{3}), -2(0) - 1 \right) ]

[ a(2) = \left( \frac{\sqrt{3}}{2} - \pi, -1 \right) ]

Therefore, at ( t = 2 ), the object's acceleration vector is ( ( \frac{\sqrt{3}}{2} - \pi, -1 ) ). To find the rate of acceleration, we calculate the magnitude of the acceleration vector:

[ |a(2)| = \sqrt{\left(\frac{\sqrt{3}}{2} - \pi\right)^2 + (-1)^2} ]

[ |a(2)| = \sqrt{\left(\frac{3}{4} - 2\pi + \pi^2\right) + 1} ]

[ |a(2)| = \sqrt{\frac{7}{4} - 2\pi + \pi^2} ]

[ |a(2)| \approx 3.655 ]

The rate of acceleration at ( t = 2 ) is approximately ( 3.655 ) and the direction of acceleration is given by the direction of the acceleration vector ( ( \frac{\sqrt{3}}{2} - \pi, -1 ) ).

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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.

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.

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.

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.

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