1.1 A fish swimming in a horizontal plane has velocity v (4.00 i +5.00 j) m/s at

Question

1.1

A fish swimming in a horizontal plane has velocity v (4.00 i +5.00 j) m/s at a point in the ocean where the position relative to a certain rock is r(-10.0 i 4.00 j) m. After the fish swims with constant acceleration for 17.0 s, its velocity is v : (15.0-5.00j) m/s. (a) What are the components of the acceleration? What is the direction of the acceleration with respect to unit vector i? ax = 0 ay = .706 m/s 0 81.95 o (counterclockwise from the +x-axis is positive) (b) If the fish maintains constant acceleration, where is it at t 26.0 s, with respect to the rock? x= (c) In what direction is it moving? (Hint: This is NOT asking for the direction of the vector you found in part b. You need to find the velocity vector.) ° (counterclockwise from the +x-axis is positive)

Explanation / Answer

[1.1]

(a)

a = (v-vi)/t = [(16.0 i - 5.00 j) - (4.00 i + 5.00 j)]/17 = [12i - 10j]/17

ax = 12.0/17.0 = 0.706 m/s2

ay = -10.0/17.0 = 0.588 m/s2

Direction:

tan^-1(-10/12) = -39.8°

-39.8° + 360° = 320.19°

(b) r = ri + (vi)t + (1/2)at^2

= (-10.0 i - 4.00 j) + 26(4.00 i + 5.00 j) + (26.0^2)(1/2)[12.0i - 10.0j]/17

= (-10.0 i - 4.00 j) + (104.00 i + 130.00 j) + (676)[6.0i - 5.0j]/17

= (94.00 i + 126.00 j) + [238.58i – 198.82j]

= (332.58i - 72.82j) m

x = 332.58 m

y = -72.82 m

(c) v = vi + at = (4.00 i + 5.00 j) + 26[12i - 10j]/17

= (22.35i – 10.29j) m/s

Direction:

tan^-1(-10.29/22.35) = -24.73°

-24.73° + 360° = 335.27°

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