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

Question

A fish swimming in a horizontal plane has velocity v = (4.00 i + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is r = (16.0 i - 1.80 j) m. After the fish swims with constant acceleration for 15.0 s, its velocity is v = (15.0 i - 6.00 j) m/s.
(a) What are the components of the acceleration of the fish?
ax = ______ m/s2
ay = ______ m/s2

(b) What is the direction of its acceleration with respect to unit vector i?
________° counterclockwise from the +x-axis

(c) If the fish maintains constant acceleration, where is it at t = 26.0 s?
x = ______ m
y = ______ m

In what direction is it moving?
________° counterclockwise from the +x-axis

Explanation / Answer

initial velocity v0 = (4.00 i + 1.00 j) m/s initial position r0 = (16.0 i - 1.80 j) m. t = 15.0 s, final velocity v = (15.0 i - 6.00 j) m/s. (a) What are the components of the acceleration of the fish? v = v0 + at a = (v - v0)/t = [(15.0 i - 6.00 j) - (4.00 i + 1.00 j)]/15.0 = (0.733 i - 0.333 j) m/s^2 so ax = 0.733 m/s2 ay = -0.333 m/s2 (b) What is the direction of its acceleration with respect to unit vector i? note ax > 0, ay < 0, the angle is in the 4th quadrant, = 360 - arctan(0.333/0.733) = 336° counterclockwise from the +x-axis (c) If the fish maintains constant acceleration, where is it at t = 26.0 s? r = r0 + v0*t + a*t^2/2 = (16.0 i - 1.80 j) + (4.00 i + 1.00 j)*26.0 + (0.733 i - 0.333 j)*26.0^2/2 = (368 i - 88.4 j) m so x = 368 m y = -88.4 m In what direction is it moving? v = v0 + a*t = (4.00 i + 1.00 j) + (0.733 i - 0.333 j)*26.0 = (23.06 i - 7.658 j) m/s angle = 360 - arctan(7.658/23.06) = 342° counterclockwise from the +x-axis

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