A fish swimming in a horizontal plane has velocity Vi = (4.00i + 1.00j ) m/s at

Question

A fish swimming in a horizontal plane has velocity Vi = (4.00i + 1.00j ) m/s at a point in the ocean where the position relative to a certain rock is ri = (11.0i ? 2.70j ) m. After the fish swims with constant acceleration for 18.0 s, its velocity is = (18.0i ? 2.00j ) m/s. The final location of the fish in unit-vector notion is the position vector rf = (454.055555555556i + -43.78 j)m. So it has coordinates (x, y) = (454 m,-43m) Its direction of motion is the direction of its velocity. Be sure that your final answer for ? is measured counterclockwise from the +x axis. ? = tan?1 (Vy/Vx)m/s (Vy/Vx)???? degree???

Explanation / Answer

cant follow the ? in the velocity and r vectors

BUT still you can figure it out this way:

let initial position vector of fish be : x1 i + y1 j

let initial velocity vector of fish be : u1 i + v1 j

let the constant acceleration be ax i + ay j

then after 18s , u2 i + v2 j = u1 i + v1 j + 18(ax i + ay j) = (u1+18ax)i + (v1+18ay)j .....(i)

let location at this instant be x2 i + y2j

then using s=ut+0.5at^2

(x2i + y2j)-(x1i + y1j) = 18( u1 i + v1 j) + 0.5(ax i + ay j)*18^2

final velocity,location seems to be given therefore find aceleration,magnitude and direction from (i)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.