Answer :
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have:
theta = arcsin(3/20) = approx. 8.63°
The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line.
x = sqrt(20^2 - 3^2)
= sqrt(400 - 9)
= sqrt 391
The boat's crossing time =
0.5 km/(sqrt 391 km/hr)
= (0.5/sqrt 391) hr
= approx. 0.025 hr
= approx. 91 seconds
theta = arcsin(3/20) = approx. 8.63°
The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line.
x = sqrt(20^2 - 3^2)
= sqrt(400 - 9)
= sqrt 391
The boat's crossing time =
0.5 km/(sqrt 391 km/hr)
= (0.5/sqrt 391) hr
= approx. 0.025 hr
= approx. 91 seconds
The time taken by the motorboat to cross the river will be t=91 seconds
What will be the time?
First we will construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = [tex]\theta[/tex] We have:
[tex]\theta=Sin^{-1}(\dfrac{3}{20})=8.63^o[/tex]
The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line.
[tex]x=\sqrt{20^2-3^2}[/tex]
[tex]x=\sqrt{400-9}[/tex]
[tex]x=\sqrt{391}[/tex]
now the boat's crossing time =
[tex]\dfrac{0.5}{\sqrt{391}}[/tex]
[tex]=0.025\ hr[/tex]
[tex]=91\ sec[/tex]
Thus the time taken by the motorboat to cross the river will be t=91 seconds
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