A fish swimming in a horizontal plane has velocity [v with arrow] i = (4.00 [i]

Question

A fish swimming in a horizontal plane has velocity [v with arrow] i = (4.00 [i] + 1.00 [j] ) m/s at a point in the ocean where the position relative to a certain rock is [r with arrow] i = (12.0 [i] 2.20 [j] ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is [v with arrow] = (19.0 [i] 3.00 [j] ) m/s. (a) What are the components of the acceleration of the fish?

ax = m/s2

ay = m/s2

(b) What is the direction of its acceleration with respect to unit vector [i] ?
° counterclockwise from the +x-axis

(c) If the fish maintains constant acceleration, where is it at t = 28.0 s?

x = m

y = m

In what direction is it moving?
° counterclockwise from the +x-axis

Explanation / Answer

Vintiial = 4i + 1 j

final velocity Vf = 19 i - 3j

time taken t = 15 secs

so

total accleration a = (vfx-vix)/t i (vfy -Viy)/t

a = (19-4) i/15 + (-3 -1) j/15

a = 1i - 0.2667 j

so ax = 1i

ay = -0.2667 j

b. direction tan theta = ay/ax = -0.2667/1 = -0.2667

theta = -14.93 deg

-------------------------------

use S = 0.5 at^2

x = 0.5 * 1 * 28 *28 = 392 m i

y = 0.5* 0.2666 *28*28 = 104.5 m j

angle = tan ^-1(104.55/392)

theta = -14.94 deg

or it ( 180-14.94) = 165 deg deom +xaxis

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