Question

The speed(v) and time(t) for an object moving along straight line are related as
t^2+100=4vt where v is in m/s and t is in seconds.Find the possible values of v.

explain the complete solution of my question step by step.

Answer 1

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'(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.