An object's two dimensional velocity is given by #v(t) = ( e^t-2t , t-4e^2 )#. What is the object's rate and direction of acceleration at #t=5 #?

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Answer 1 · 2018-03-07T16:49:22

The rate of acceleration is #=146.4ms^-1# in the direction of #0.4^@# anticlockwise from the x-axis

The acceleration is the derivative of the velocity

#v(t)=(e^t-2t, t-4e^2)#

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

When #t=5#

#a(5)=(e^5-2, 1)#

#a(5)=(146.4,1)#

The rate of acceleration is

#||a(5)||=sqrt(146.4^2+1^2)=146.4ms^-1#

The angle is

#theta=arctan(1/146.4)=0.4^@# anticlockwise from the x-axis

Answer 2 · 2018-03-07T16:50:17

Please see below.

As velocity is given by #v(t)=(e^t-2t,t-4e^2)#

accelaration is #(dv)/(dt)=d/(dt)(e^t-2t)hati+d/(dt)(t-4e^2)hatj#

or #(e^t-2)hati+hatj#

Note that #-4e^2# is a constant hence its differential is #0#.

and at #t=5#, accelaration is #(e^5-2)hati+hatj#

Its magnitude is #sqrt((e^5-2)^2+1)=sqrt((148.4132-2)^2+1)#

= #146.417#

and direction is #tan^(-1)(1/(e^5-2))=tan^(-1)(1/146.4132)#

= #0.39^@#