A fish swimming in a horizontal plane has velocity v with arrowi = (4.00 i + 1.0

Question

A fish swimming in a horizontal plane has velocity v with arrowi = (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 arrowi = (15.0 i 1.70 j) m. After the fish swims with constant acceleration for 16.0 s, its velocity is v with arrow = (22.0 i 2.00 j) m/s. (a) What are the components of the acceleration of the fish? (b) What is the direction of its acceleration with respect to unit vector i? (c) If the fish maintains constant acceleration, where is it at t = 30.0 s? In what direction is it moving?

Explanation / Answer

(a)
a = (v-vi)/t = [(15.0 i - 1.70 j) - (4.00 i + 1.00 j)]/16 = [11i - 2.70j]/16
ax = 11.0/16.0 m/s2 =0.6875 m/s^2
ay = -2.70/16.0 m/s2 = -0.168m/s^2

(b)
tan^-1(-2.7/11) = -13.79°
-13.79° + 360° = 346.21°

(c) r = ri + (vi)t + (1/2)at^2
= (15.0 i - 1.700 j) + 30(4.00 i + 1.00 j) + (30.0^2)(1/2)[18.0i - 3.0j]/16
= (10.0 i - 4.00 j) + (120.00 i + 30.00 j) + (900)[18.0i - 3.0j]/16
= (130.00 i + 26.00 j) + [1012.5i - 168.75j]
= (1140.5i - 142.75j) m
x = 1140.5 m
y = -142.75 m

v = vi + at = (4.00 i + 1.00 j) + 30[18i - 3j]/16
= (37.5i - 6.625j) m/s
tan^-1(-6.625/37.5) = -10.01°
-10.01° + 360° = 349.99°

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