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## Homework Statement

A fish swimming in a horizontal plane has a velocity of 4.123 m/s [N 75.96deg E] in the ocean with a position vector of 10.770 m [S 68.20 deg E] relative to a stationary rock at the shore. After the fish swims with constant acceleration for 20.00 s, its velocity is 20.62 m/s [S 75.96 deg E].

(a) Calculate the acceleration.

(b) Where is the fish at 25.00 s and in what direction is it moving?

## Homework Equations

A

_{T}= A

_{X}+ A

_{Y}

N, E (+)

V

_{2x}= -(sin75.96)(20.62)

V

_{1x}= (cos14.04)(4.123)

V

_{2y}= (cos75.96)(20.62)

V

_{1y}= (sin14.04)(4.123)

## The Attempt at a Solution

(a)

x-dir

A

_{x}= (-V

_{2x}- V

_{1x}) / t

= ( -(sin75.96)(20.62) - (cos14.04)(4.123) ) / 20

= -1.2002 m/s

^{2}

y-dir

A

_{y}= (V

_{2y}- V

_{1y}) / t

= ( (cos75.96)(20.62) - (sin14.04)(4.123) ) / 20

= 0.2001 m/s

^{2}

A

_{t}= sqrt(a

_{x}

^{2}+ a

_{y}

^{2})

= 1.217 m/s

^{2}

tan(theta) = ax/ay

theta = 81deg

Therefore a

_{T}= 1.217 m/s

^{2}[S 81deg E]

(b) For b, I know I would split up D

_{T}into two components the same way I did above.

I would use the ax, ay, t=25, v1x, v1y, d1x and d1y that I already have

For example,

dx = v1x (t) + 1/2 at^2

dx = (cos14.04)(4.123) + 1/2(-1.2001)^2

This is asking the distance at 25s.. The problem states that the position vector is 10.770m[ S 68.26 E] meaning that I would have to add this to the equation, I think.

I would take the x-component of the position vector (sin68.20)(10.770) and add it to the dx equation I wrote above, like so:

dx = (sin68.20(10.770) + (cos14.04)(4.123) + 1/2(-1.2001)^2

then do the same for dy and then find the dt value + magnitude.

**Did I do both parts correctly?**

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