A fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is i = (16.0 î − 3.20 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (21.0 î − 6.00 ĵ) m/s. In what direction is it moving? ______________? ° counterclockwise from the +x-axis
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A fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is i = (16.0 î − 3.20 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (21.0 î − 6.00 ĵ) m/s.
In what direction is it moving?
______________? ° counterclockwise from the +x-axis
Acceleration of the fish is
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- A fish swimming in a horizontal plane has velocity v = (4.00 î + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is ï = (10.0 î – 4.00 j) m . After the fish swims with constant acceleration for 20.0 s, its velocity is v = (20.0 î – 5.00 j) m/s . 1. What are the components of the acceleration of the fish? 2. What is the direction of its acceleration with respect to unit vector î ? 3. If the fish maintains constant acceleration, where is it at t = 25.0 s and in what direction is it moving?A fish swimming in a horizontal plane has velocity given by v i j i = 4ˆ + ˆ m/s at a point in theocean where the position relative to a certain rock is r i j i = 10ˆ − 4 ˆ m. After the fish swimswith constant acceleration for 20.0 s, its velocity is v i j = 20ˆ − 5 ˆ m/s.a) What are the components of the acceleration of the fish?b) What is the direction of its acceleration with respect to the unit vector along the x axis?c) If the fish maintains constant acceleration, where is it at t = 25.0 s and in what direction isit moving?A particle leaves the origin with initial velocity v0= 12i + 14j m/s, undergoing constant acceleration a = -0.80i + 0.25j m/s2. If the particle crosses the y-axis at t = 30s and its y-coordinate at the time is 532.5m. How fast is it moving and in what direction is it moving?
- A fish swimming in a horizontal place has velocity v = (4.01 +1.0))m/s at a point in the ocean where the position relative to a rock is = (10.0-4.0)m. After the fish swims with constant acceleration for 20 sec its velocity is v = (20.01-5.0/)m/s. A) What are the components of the acceleration? B) What is the direction of the acceleration with respect to unit vector ? C) If the fish maintains constant acceleration, where is it at t=2.5 s, and in what direction is it moving? 4) A stone is thrown straightA fish swimming in a horizontal plane has velocity V₁ = (4.00 i + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is r = (16.01- 3.60 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (15.01 - 1.00 j) m/s. (a) What are the components of the acceleration of the fish? ax = m/s2 m/s2 ay = (b) What is the direction of its acceleration with respect to unit vector î? counterclockwise from the +x-axis (c) If the fish maintains constant acceleration, where is it at t = 24.0 s? x = y = m In what direction is it moving? counterclockwise from the +x-axisA fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is i = (12.0 î − 3.60 ĵ) m. After the fish swims with constant acceleration for 17.0 s, its velocity is = (19.0 î − 6.00 ĵ) m/s.
- A fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is i = (12.0 î − 2.60 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (25.0 î − 1.00 ĵ) 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 î? Draw coordinate axes on a separate piece of paper, and then add the acceleration vector with its tail at the origin. Write the numerical values for the x and y components and then use this drawing to determine the angle.° counterclockwise from the +x-axis(c) If the fish maintains constant acceleration, where is it at t = 28.0 s? x = m y = m In what direction is it moving? ° counterclockwise from the +x-axisA fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is i = (16.0 î − 3.20 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (21.0 î − 6.00 ĵ) 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 î? ° counterclockwise from the +x-axis(c) If the fish maintains constant acceleration, where is it at t = 30.0 s? x = m y = m In what direction is it moving? ° counterclockwise from the +x-axisA fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative to a certain rock is i = (16.0 î − 3.20 ĵ) m. After the fish swims with constant acceleration for 15.0 s, its velocity is = (21.0 î − 6.00 ĵ) 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 î? ° counterclockwise from the +x-axis(c) If the fish maintains constant acceleration, where is it at t = 30.0 s? x = 542.5 Remember that you can treat the motions in the x and y directions separately. Each is then treated exactly as you would the one-dimensional case. m y = m In what direction is it moving? ° counterclockwise from the +x-axi
- An ion's position vector is initially 7 = (-3,9 m )i + (-8.6 m )j + (-3.4 m )k, and 9.3 s later it is 7 = (-2.6 m )i + (7.3 m )ĵ + (1.0 m )R. In unit-vector notation, what is its average velocity during the 9.3 s? Number i k Units m/sA car moves along a horizonatal road with constant velocityv0 = v0, x^ii = (30.06 m/s)^i until it encounters a smooth inclined hill. It climbs the hill with constant velocity. v1=v1x^+v1y^j = (44.2 m/s)^i + (3.23 m/s)^jWhat is the expression in cartesian unit vector notation, for average acceleration of the car during the given time period using the symbols provided?A cannon ball is fired with an initial speed of 123 m/s at angle of 60 degrees from the horizontal. Express the initial velocity as a linear combination of its unit vector components. Vo = ( m/s) î + ( /s) ? m At the maximum height, the speed of the cannon ball is v = m/s and the magnitude of its acceleration is a = m/s?. The time needed to reach maximum height ist= S. The maximum height reached by the cannon ball is H = m.