A fish swimming in a horizontal plane has velocity Vi (4.o0i1.00 j) m/s at a poi

Question

A fish swimming in a horizontal plane has velocity Vi (4.o0i1.00 j) m/s at a point in the ocean where the position relative to a certain rock is r velocity is v = (25.0 i-1.00 j) m/s. (16.0 i-1.20 j) m. After the fish swims with constant acceleration for 17.0 s, its (a) What are the components of the acceleration of the fish? ax-1.24 m/s2 a,-.0.12 (b) What is the direction of its acceleration with respect to unit vector ? -5.44 counterclockwise from the +x-axis (c) If the fish maintains constant acceleration, where is it at t = 28.0 s? x-614.08 y=-19.04 In what direction is it moving? 1.78 Draw coordinate axes on a separate piece of paper, and then add the velocity vector with its tail at the origin. Write the numerical values for the x and y components and then use this drawing to determine the angle. counterclockwise from the +x-axis

Explanation / Answer

a] By using equation of motion,

v = vi + at

a = v - vi /t

= [(25i - 1j) - (4i + 1j)]/17

= [21i -2j]/17

ax = 21/17 = 1.24 m/s^2

ay = -2/17 = -0.12 m/s^2

b] arctan(-2/21) = -5.44 degre counterclockwise from the +x-axis

c] r = ri + vi*t + 0.5at^2

= (16i - 1.20j) + (4i + 1j)*28 + 0.5*(21i - 2j)*(28)^2/17

= (16 + 4*28 + 0.5*21*28^2/17)i + (-1.20 + 28 -0.5*2*28^2)j

= 612.35i - 19.32j

x = 612.35 m

y = 19.32 m

d] v = vi + at

= (4i + 1j) + (21i - 2j)*28/17

= (4 + 21*28/17)i + (1 - 2*28/17)j

= 38.59i - 2.29j

arctan(-2.29/38.59) = -3.396 degree

answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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