A swimmer, capable of swimming at a speed of 1.60 m/s in still water (i. e., the swimmer can swim with a speed of 1.60 m/s relative to the water), starts to swim directly across a 1.25-km-wide river. However, the current is 0.549 m/s, and it carries the swimmer downstream. (a) How long does it take the swimmer to cross the river

  t = 781.25 s

Explanation:

This is an exercise in velocity composition, if we set a reference system where the x-axis is perpendicular to the river and the y-axis is parallel to the river.

The swimmer has a velocity on the x axis

           vx = 1.60 m / s

a velocity on the y axis, created by the current of the river

           vy = 0.549 m / s

time is a scalar, therefore the time it takes to cross the river is the same time it creates the displacement in e; Axis y

X axis

            vₓ = x / t

            t = x / vₓ

            t = 1250 / 1.6

            t = 781.25 s

in this time a distance has descended

            y = v_y t

            y = 0.549 781.25

            y = 428.9 m

this could be newton#'s law of cooling. the cooling curve is exponential

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft