Answer:
9.51
Explanation:
The distance s is given by:
[tex]s(t) = \sqrt{151^2 + (vt)^2}[/tex]
The change in distance is given by the time derivative of s:
[tex]\frac{ds}{dt} = \frac{v^2t}{\sqrt{151^2 + (vt)^2}}[/tex]
For the time t you solve the equation of distance x for time:
[tex]x = vt => t = \frac{x}{v}[/tex]
Plugging in for t:
[tex]\frac{ds}{dt}(t=\frac{x}{v})=\frac{vx }{\sqrt{151^2 + x^2}}=9.51[/tex]
