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

Question

A fish swimming in a horizontal plane has velocity i = (4.00 + 1.00 ) m/s at a point in the ocean where the position relative to a certain rock is i = (12.0 ? 1.80 ) m. After the fish swims with constant acceleration for 19.0 s, its velocity is = (17.0 ? 1.00 ) m/s.
(a) What are the components of the acceleration of the fish?
ax =

m/s2
ay = -0.10


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 = [(20.0 i - 5.00 j) - (4.00 i + 1.00 j)]/23 = [16i - 6j]/23 ax = 16.0/23.0 m/s2 ay = -6.0/23.0 m/s2 (b) tan^-1(-6/16) = -20.56° -20.56° + 360° = 339.44° (c) r = ri + (vi)t + (1/2)at^2 = (10.0 i - 4.00 j) + 25(4.00 i + 1.00 j) + (25.0^2)(1/2)[16.0i - 6.0j]/23 = (10.0 i - 4.00 j) + (100.00 i + 25.00 j) + (625)[8.0i - 3.0j]/23 = (110.00 i + 21.00 j) + [217i - 81.5j] = (327i - 60.5j) m x = 327 m y = -60.5 m v = vi + at = (4.00 i + 1.00 j) + 25[16i - 6j]/23 = (21.4i - 5.5j) m/s tan^-1(-5.5/21.4) = -14.5° -14.5° + 360° = 345.5°

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