1. 2.4/12 points | Previous Answers My Notes A fish swimming in a horizontal pla
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1. 2.4/12 points | Previous Answers My Notes A fish swimming in a horizontal plane has velocity vi = (4.00 i-3.00 j) m/s at a point in the ocean where the position relative to a certain rock is ri (-10.0 i 4.00 j) m. After the fish swims with constant acceleration for 21.0 s, its velocity is v = (23.0 i -5.00 j) m/s. (a) What are the components of the acceleration? What is the direction of the acceleration with respect to unit vector i? X- | 0.9 m/s^2 Enter a number. [S71-0 x (counterclockwise from the +x-axis is positive) 26.0 s, (b) If the fish maintains constant acceleration, where is it at t with respect to the rock? (c) In what direction is it moving? (Hint: This is NOT asking for the direction of the vector you found in part b. You need to find the velocity vector.) × (counterclockwise from the +x-axis is positive)Explanation / Answer
a) initial velocity in x direction = 4 m/s
final velocity in x direction = 23 m/s
acceleration = (23-4)/21=0.905 m/s2
initial velocity in y direction = -3 m/s
final velocity in y direction = -5 m/s
acceleration = [-5-(-3)]/21=-0.095 m/s2
vectorially, acceleration = 0.905i-0.095j m/s2
direction of acceleration measured counteclockwise from +x axis = 360-tan-1(.095/0.905)=354o
b) distnace travelled in x direction = 4*26 + 0.5*0.905*262 = 409.89 m
distnace travelled in negative y direction = 3*26 +0.5* 0.095*262 = 110.11 m
position of fish with respect to rock = (-10+409.89)i+(-4-110.11)j=399.89i-114.11j
x=399.89m
y=-114.11 m
c)velocity in x direction after 26 seconds = 4+0.905*26=27.53 m/s
velocity in y direction after 26 seconds = -3-0.095*26=-5.47 m/s
velocity in vector form= 27.53i-5.47 j m/s
direction measured counterclockwise from +x axis = 360-tan-1(5.47/27.53)=348.76o
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