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

Answer 1

Elena Albarran

The rate of acceleration is #=sqrt5ms^-2# in
the direction is #=26.6^@# clockwise from the x-axis.

Acceleration is the velocity derivative.

#v(t)=(t^2-2t, cospit-t)#

#a(t)=v'(t)=(2t-2, -pisinpit-1)#

When #t=2#

#a(2)=v'(2)=(2*2-2, -pisinpi*2-1)#

#=(2,-1)#

The acceleration rate is

#||a(2)||=||(2,-1)||=sqrt(2^2+(-1)^2)=sqrt5 ms^-2#

The path is

#theta=arctan(-1/2)=26.6^@# clockwise from the x-axis.

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Answer 2

Carson Alexander

To find the acceleration, differentiate the velocity function with respect to time twice. Then evaluate the resulting expression at ( t = 2 ).

First derivative of velocity: ( v'(t) = (2t - 2, -\sin(pt) \cdot p - 1) )

Second derivative of velocity: ( a(t) = v''(t) = (2, -\cos(pt) \cdot p^2) )

At ( t = 2 ), the acceleration vector is: ( a(2) = (2, -\cos(2p) \cdot p^2) )

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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.