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(a) In which direction should the motorb...

(a) In which direction should the motorboat given in Illustration `5.46` head in order to reach a point on the opposite bank directly east from the starting point ? The boat's speed relative to the water remains `4 ms^-1`.
(b) What is the velocity of the boat relative to the earth ?
( c) How much time is required to cross the river ?

Text Solution

Verified by Experts

As the boat has to each exact opposite end to the point of start, the boat has to start (velocity `4 m s^-1`) at an angle `theta` aiming somewhat upstream. Taking into count the push given by the current,
Velocity of boat w.r.t. river, `vec v_(b,r) = vec(OA) = 4 m s^-1`
Velocity of river w.r.t. Earth, `vec v_r = vec(AB) = 2 m s^-1`
velocity of boat w.r.t. Earth, `vec v_b m s^-1 = vec(OB) = ?`
(a) To find the direction of the boat in which the boat has to go,we need to find angle `theta`.
From `Delta OBA, sin theta = (AB)/(OA) = (2)/(4) = (1)/(2) rArr theta = 30^@`
Hence, the motorboat has to head at `30^@` north of east.
(b) To find the velocity of boat w.r.t. Earth, we can use pythagorous theorem again.
Fro `Delta OBA`, we have
`v_(b,r)^2 = v_b^2 + v_r^2 rArr v_b^2 = v_(b,r)^2 - v_r^2`
`rArr v_b^2 = sqrt(v_(b,r)^2 - v_r^2) = sqrt(4^2 - 2^2) = 2 sqrt(3) m s^-1`
( c) Time taken to cross the river is
(Width of river)/(Velocity of boat w.r.t Earth)
`rArr t = (800)/(2 sqrt(3)) = (400 sqrt(3))/(3) s`.
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