A fish swimming in a horizontal plane has velocity v i = (4.00 i - 5.00 j ) m/s

Question

A fish swimming in a horizontal plane has velocityvi=(4.00 i - 5.00 j) m/s at a point in theocean where the position relative to a certain rock isri=(10.0 i - 4.00 j) m. After the fish swimswith constant acceleration for 15.0 s,its velocity is v = (18.0 i -5.00 j) m/s. (a) What are the components of theacceleration?
ax = 1m/s2
ay = 2m/s2

(b) What is the direction of the acceleration with respect to unitvector i?
3° (counterclockwise from the+x-axis is positive)

(c) If the fish maintains constant acceleration, where is it att = 22.0 s?
x = 4 m
y = 5 m
In what direction is it moving?
6° (counterclockwise from the+x-axis is positive) (a) What are the components of theacceleration?
ax = 1m/s2
ay = 2m/s2

(b) What is the direction of the acceleration with respect to unitvector i?
3° (counterclockwise from the+x-axis is positive)

(c) If the fish maintains constant acceleration, where is it att = 22.0 s?
x = 4 m
y = 5 m
In what direction is it moving?
6° (counterclockwise from the+x-axis is positive)

Explanation / Answer

(a)We know from the relation vf = vi + at or a = (vf - vi/t) vf = (18.0i - 5.00j) m/s,vi = (4.00 i -5.00 j) m/s and t = 15.0 s (b)The direction of the acceleration with respect to unitvector i tan = (ay/ax) or = tan-1(ay/ax)(counterclockwise from the +x-axis is positive) (c)When t = 22.0 s then r = vit + (1/2)at2 The direction in which it is moving is tan = (ry/rx) or = tan-1(ry/rx)(counterclockwise from the +x-axis is positive)

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