show work pls 12. 0.48/1 points| Previous Answers SerPSE10 4.2.OP.005.MI. A fish
ID: 1885810 • Letter: S
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show work pls
12. 0.48/1 points| Previous Answers SerPSE10 4.2.OP.005.MI. A fish swimming in a horizontal plane has velocity V: = (4.00 + 1.00 j) m/s at a point in the ocean where the position relative to a certain rock is r = (12.0 i-3.60 j) m. After the fish swims with constant acceleration for 15.0 s, its velocity is v-(21.0 i-3.00 j) m/s. My Notes Ask Your Teacher (a) What are the components of the acceleration of the fish? a17/15 3, =1-4/15 m/s (b) What is the direction of its acceleration with respect to unit vector i 13.24 counterclockwise from the tx-axis (c) If the fish maintains constant acceleration, where is it at t-26.0 s? 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 It seems you have forgotten that the fish had an initial velocity at time zero. m In what direction is it moving? ocounterclockwise from the +x-axis Need Help? Read tMaster Master It a Show My Work (opionan 13. -1 points SerPSE10 4.2.P.005 A snowmobile is originally at the point with position vector 31.5 m at 95.0° counterclockwise from the x axis, moving with velocity 4.85 m/s at 40.0°. It moves with constant acceleration 1.91 m/s at 200°. After 5.00 s have elapsed, find the following. My Notes Ask Your Teacher (a) its velocity vector m/s (b) its position vector Need Help? Read It a Show My Work (opionanExplanation / Answer
2.
vi = initial velocity of taxi = 34 mi/h = 49.87 ft/s
vf = final velocity of taxi = 0 ft/s
d = stopping distance for taxi = 50 ft
a = acceleration
using the equation
vf2 = vi2 + 2 a d
02 = 49.872 + 2 a (50)
a = - 24.87 m/s2
when
vi = initial velocity of taxi = 69 mi/h = 101.2 ft/s
vf = final velocity of taxi = 0 ft/s
d = stopping distance for taxi = ?
a = acceleration = - 24.87 m/s2
using the equation
vf2 = vi2 + 2 a d
02 = 101.22 + 2 (- 24.87) d
d = 205.89 ft
13.
ri = initial position = 31.5 Cos95 i + 31.5 Sin95 j = - 2.75 i + 31.4 j
vi = initial velocity = (4.85 Cos40) i + (4.85 Sin40) j = 3.72 i + 3.12 j
a = acceleration = (1.91 Cos200) i + (1.91 Sin200) j = - 1.79 i - 0.65 j
t = time taken = 5 sec
vf = final velocity = ?
Using the equation
vf = vi + a t
vf = (3.72 i + 3.12 j) + (- 1.79 i - 0.65 j) (5)
vf = - 5.23 i - 0.13 j
magnitude : sqrt((- 5.23)2 + (- 0.13)2) = 5.23 m/s
direction : 180 + tan-1(0.13/5.23) = 181.43 deg
b)
final position is given as
rf = ri + vi t + (0.5) a t2
rf = (- 2.75 i + 31.4 j ) + (3.72 i + 3.12 j) (5) + (0.5) (- 1.79 i - 0.65 j) (5)2
rf = (- 2.75 i + 31.4 j ) + (18.6 i + 15.6 j) + (- 22.375 i - 8.125)
rf = - 6.525 i + 38.875 j
magnitude : sqrt((- 6.525)2 + (38.875)2) = 39.4 m
direction : 180 - tan-1(38.875/6.525) = 99.5 deg
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