A fish swimming in a horizontal plane has velocity vi = (4.00 i + 5.00 j) m/s at

Question

A fish swimming in a horizontal plane has velocity vi = (4.00 i + 5.00 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 17.0 s, its velocity is v = (18.0 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?

ax = ?

ay = ?

= ? ° (counterclockwise from the +x-axis is positive)

(b) If the fish maintains constant acceleration, where is it at t = 26.0 s, with respect to the rock?

x = ?

y = ?

(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

here,

initial speed of the fish , vi = ( 4i +5 j ) m/s

ri = ( - 10 i - 4 j) m

final velocity ,v = ( 18 i - 5 j) m/s

time taken , t = 17 s

(a)

accelration , a = ( v - vi)/t

a = ( 14 i - 10 j) /( 17)

a = ( 0.82 i - 0.59 j ) m/s^2

ax = 0.82 m/s^2

ay = - 0.59 m/s^2

theta = arctan( 0.59/0.82)

theta = 35.74 degree

the angle theta is 35.74 degree clockwise from the + x axis

(b)

when t = 26 s

x = x0 + vi*t + 0.5 *ax*t^2

x = - 10 + 4 * 26 + 0.5 * 0.82* 26^2

x = 371.16 i m

y = y0 + vi*t + 0.5 *ay*t^2

y = - 4 +5 * 26 - 0.5 * 0.59 * 26^2

y = - 73.42 m

(c)

the velocity at t = 26 s,

v'x = ( vix + ax*t )

v'x = 4 + 0.82 * 26

v'x = 25.31 m/s

vy' = ( viy + ay * t)

vy' = 5 - 0.59 * 26

vy' = - 10.34 m/s

theta = arctan( 10.34 / 25.31)

theta = 22.22 degree

the angle theta is - 22.22 degree

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