if t^2+100=4vt then find the possible values of v

mechanics

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  • We solve this equation for v:

    4vt=100+t^2, v=(100+t^2)/4t.

    Now, to the possible values.

    We cant calculate v when t=0 (divizion by 0). Also, t can not be negative, so v too. Then we find the minimum for v(t).



    v^2;(t)=-25/t^2+1/4=0 so t=10 is a local minimum.

    Minimal value for v is v(10)=(100+100)/40=5m/s maximum is any.

    So the answer - v is equal or greater than 5 m/s.
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