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

Question

A fish swimming in a horizontal plane has velocity Vi = (4.00 + 1.00 ) m/s at a point in the ocean where the position relative to a certain rock is ri = (16.0 - 2.20 ) m. After the fish swims with constant acceleration for 19.0 s, its velocity is V= (25.0 ? 1.00 ) m/s.
(a) What are the components of the acceleration of the fish?
ax =_________________m/s2
ay =_________________m/s2

Review the definition of average acceleration and remember that each component is treated separately. _____________________m/s2


Review the definition of average acceleration and remember that each component is treated separately. _____________________m/s2

(b) What is the direction of its acceleration with respect to unit vector ?
___________________

Explanation / Answer

(a) a = (v-vi)/t = [(25.0 i - 1.00 j) - (4.00 i + 1.00 j)]/19= [21i - 2j]/19 ax = 21.0/19.0 m/s2 ay = -2.0/19.0 m/s2 (b) tan^-1(-2/21) = -5.44° -5.44° + 360° = 354.55 degrees (c) r = ri + (vi)t + (1/2)at^2 = (16.0 i - 2.20 j) + 24(4.00 i + 1.00 j) + (25.0^2)(1/2)[21.0i - 2.0j]/19 = (16.0 i - 2.20 j) + (96.00 i + 24.00 j) + (625)[10.5i - 1.5.0j]/19 = (112.00 i + 21.80 j) + [345.72i - 49.33j] = (457.72i - 71.13j) m x = 457.72 m y = -71.3 m v = vi + at = (4.00 i + 1.00 j) + 24[21i - 2j]/19 = (30.52i - 1.52j) m/s tan^-1(-1.52/30.52) = -2.86° -2.86° + 360° = 357.13°

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