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What is the instantaneous velocity of an object moving in accordance to # f(t)= (sin2t-cos2t,sin(2t-pi/4)) # at # t=(-pi)/3 #?

Answer 1

#v#=#3.346# length unit/unit of time
And makes an angle with the positive direction of #x# axis #theta #
#theta=190.72# degrees

This function is a Parametric function of motion of a particle which represents the relation between #x,y# and #t#. #f(t)=(x(t),y(t))# By finding the first derivative #f'(t)=(x'(t),y'(t))#

#f'(t)=(2cos2t+2sin2t,2cos(2t-pi/4))#

substitute for #t=-pi/3#

the horizontal component of the velocity of the particle#(y^0)# =#-1-sqrt3#

the vertical component of the velocity of particle#(x^0)# =#(-sqrt6+sqrt2)/2#

so the magnitude of the velocity of the particle=#sqrt((x^0)^2+(y^0)^2#

=#3.346# length unit/unit of time

and its direction can be given through the relation

#tantheta=y^0/x^0#=#0.18946#

#theta=10.72# degrees

but it's in the third quadrant since both #x^0,y^0# are negative values so #theta=190.72# degrees

I hope this was helpful.

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Answer 2

Paisley Johnson

To find the instantaneous velocity of an object moving according to the function ( f(t) = (\sin(2t) - \cos(2t), \sin(2t - \frac{\pi}{4})) ) at ( t = -\frac{\pi}{3} ), we first need to differentiate the function with respect to ( t ) to find the velocity vector function. Then, we substitute ( t = -\frac{\pi}{3} ) into the velocity vector function to find the instantaneous velocity at that particular time.

Taking the derivative of ( f(t) ) with respect to ( t ), we get:

[ \frac{df}{dt} = \left( \frac{d}{dt}(\sin(2t) - \cos(2t)), \frac{d}{dt}(\sin(2t - \frac{\pi}{4})) \right) ]

Using the chain rule and the derivatives of sine and cosine functions, we get:

[ \frac{df}{dt} = (2\cos(2t) + 2\sin(2t), 2\cos(2t - \frac{\pi}{4})) ]

Now, substitute ( t = -\frac{\pi}{3} ) into the velocity vector function:

[ \frac{df}{dt}\bigg|_{t = -\frac{\pi}{3}} = (2\cos(-\frac{2\pi}{3}) + 2\sin(-\frac{2\pi}{3}), 2\cos(-\frac{2\pi}{3} - \frac{\pi}{4})) ]

Simplify the trigonometric functions:

[ \frac{df}{dt}\bigg|_{t = -\frac{\pi}{3}} = (-\sqrt{3} - 2, 2\cos(-\frac{11\pi}{12})) ]

Finally, compute the cosine value:

[ \cos(-\frac{11\pi}{12}) \approx -0.2588 ]

Therefore, the instantaneous velocity of the object at ( t = -\frac{\pi}{3} ) is approximately:

[ (-\sqrt{3} - 2, 2 \times (-0.2588)) ]

[ \approx (-2.732, -0.5176) ]

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Answer from HIX Tutor

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