Question #233707

Let the position vector of a stone at time,t be given as r(t)=cosh (t^2-1)I+ sinh(1-t)j+Bt^2k

Assume that the position vector is Normal to the acceleration vector.

Find the value of B at Time t=5seconds


1
Expert's answer
2021-09-07T02:34:44-0400
r(t)=cosh(t21)i+sinh(1t)j+Bt2kr(t)=\cosh (t^2-1)i+ \sinh(1-t)j+Bt^2k

v(t)=r(t)v(t)=r'(t)

=2tsinh(t21)icosh(1t)j+2Btk=2t\sinh (t^2-1)i- \cosh(1-t)j+2Btk


a(t)=r(t)a(t)=r''(t)

=(2sinh(t21)+4t2cosh(t21))i+sinh(1t)j+2Bk=(2\sinh (t^2-1)+4t^2\cosh (t^2-1))i+ \sinh(1-t)j+2Bk

r(t)a(t)=0r(t)\cdot a(t)=0

cosh(t21)(2sinh(t21)+4t2cosh(t21))\cosh (t^2-1)(2\sinh (t^2-1)+4t^2\cosh (t^2-1))

+sinh(1t)sinh(1t)+Bt2(2B)=0+ \sinh(1-t) \sinh(1-t)+Bt^2(2B)=0

B2t2=cosh(t21)sinh(t21)B^2t^2=-\cosh (t^2-1)\sinh (t^2-1)

2t2cosh2(t21)sinh2(1t)-2t^2\cosh^2 (t^2-1)-\sinh^2(1-t)

t=5t=5

25B2=cosh(251)sinh(251)25B^2=-\cosh (25-1)\sinh (25-1)

2(25)cosh2(251)sinh2(15)-2(25)\cosh^2 (25-1)-\sinh^2(1-5)

B2=125(sinh(48)250cosh2(24)+sinh2(4))<0B^2=\dfrac{1}{25}(-\dfrac{\sinh(48)}{2}-50\cosh^2 (24)+\sinh^2(4))<0

There are no solution for BR.B\in \R.


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