Question #11265

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.

Expert's answer

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.

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.

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