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

Question

A fish swimming in a horizontal plane has velocityvi=(4.00 i + 1.00 j) m/s at a point in the oceanwhere the position relative to a certain rock isri=(10.0 i + 4.00 j) m. After the fish swimswith constant acceleration for 25.0 s, itsvelocity is v = (25.0i - 5.00j) m/s. (a) What are the components of theacceleration?
ax = 1 m/s2
ay = 2 m/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 = 35.0 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 = 1 m/s2
ay = 2 m/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 = 35.0 s?
x = 4 m
y = 5 m
In what direction is it moving?
6° (counterclockwise from the+x-axis is positive)

Explanation / Answer

The velocity of the fish in the horizontal plane isvi = (4.00i + 1.00j) m/s at a point in the ocean wherethe position relative to a certain rock is ri= (10.0i + 4.00j) m The fish swims with constant acceleration for t = 25.0 s andits velocity is v = (25.0i - 5.00j) m/s (a)Let the acceleration of the fish be a.Therefore,weget v = vi + at or a = (v - vi/t) or a = ((25.0i - 5.00j) - (4.00i + 1.00j)/(25.0)) or a = ((21.0i - 4.00j)/(25.0)) = ((21.0/25.0)i -(4.00/25.0)j) The magnitude of the acceleration is |a| = ((21.0/25.0)2 +(4.00/25.0)2)1/2 or |a| = (21.38/25) m/s2 = 0.8552m/s2 The angle between the velocity vectors is tan = (vi/v) or = tan-1(vi/v) Here,|vi| = ((4.00)1/2 +(1.00)1/2)1/2 = (17)1/2 = 4.12m/s and |v| = ((25.0)2 +(5.00)2)1/2 = 25.49 m/s or = tan-1(|vi|/|v|) =tan-1(4.12/25.49) or = 9.2o The components of the acceleration are The components of the acceleration are ax = a * cos = 0.8552 *cos(9.2o) = 0.8442 m/s2 and ay = a * sin = 0.8552 *sin(9.2o) = 0.1367 m/s2 (b)The direction of the acceleration with respect to unitvector i is = 9.2o. (c)When the fish maintains constant acceleration,the distancetravelled by the fish is S = vit + (1/2)at2 Here,t = 35.0s or S = (4.00i + 1.00j) * (35.0) + (1/2) * ((21.0/25.0)i -(4.00/25.0)j) * (35.0)2 or S = (140i + 35j) + 24.5 * (21.0i - 4.00j) = (140i + 35j) +(514.5i - 98j) or S = 654.5i - 63j Therefore,we get x = 654.5 m and y = -63 m The direction in which it is moving is tan = (y/x) = (-63/654.5) or = tan-1(-63/654.5) or S = (140i + 35j) + 24.5 * (21.0i - 4.00j) = (140i + 35j) +(514.5i - 98j) or S = 654.5i - 63j Therefore,we get x = 654.5 m and y = -63 m The direction in which it is moving is tan = (y/x) = (-63/654.5) or = tan-1(-63/654.5) or = 5.5o

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