Answer:
1.0198m/s
Explanation:
let the distance of boat from the dock be [tex]x[/tex], then use Pythagorean theorem we have [tex]x^2+1^2=y^2[/tex]: where [tex]y[/tex] the length of the rope:
#Differentiate through with respect to [tex]t[/tex]
[tex]2x.\frac{dx}{dt}=2y.\frac{dy}{dt}[/tex]
Finding [tex]\frac{dx}{dt}[/tex], and [tex]\frac{dy}{dt}[/tex] is given as [tex]1ms^-^1[/tex]
With [tex]x=5m[/tex] at the given instant and [tex]y=\sqrt{26}[/tex]
Hence
[tex]5.\frac{dx}{dt}=\sqrt{5^2+1^2}=\sqrt{26}[/tex]
Rate of decrease of the distance between dock and boat is [tex]\frac{\sqrt 26}{5}[/tex]=1.0198m/s
