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 = (14.0 ? 4.20 ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (25.0 ? 1.00 ) m/s.

(a) What are the components of the acceleration of the fish?
ax =
ay =
(b) What is the direction of its acceleration with respect to unit vector ?
(c) If the fish maintains constant acceleration, where is it at t = 28.0 s?
x =
y =
(d) In what direction is it moving?

Explanation / Answer

a) ax= 25-4 / 15 =21/15 =1.4 m/s^2 ay = (-1-1)/15 =-2/15 = -0.133 m/s^2 b)direction = tan inverse(-0.1333/1.4) = -5.44 degrees c)x=674.8 m y = -28.456 d) -2.1 degrees

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