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?
AT = AX + AY
N, E (+)
V2x = -(sin75.96)(20.62)
V1x = (cos14.04)(4.123)
V2y = (cos75.96)(20.62)
V1y = (sin14.04)(4.123)
The Attempt at a Solution
Ax = (-V2x - V1x) / t
= ( -(sin75.96)(20.62) - (cos14.04)(4.123) ) / 20
= -1.2002 m/s2
Ay = (V2y - V1y) / t
= ( (cos75.96)(20.62) - (sin14.04)(4.123) ) / 20
= 0.2001 m/s 2
At = sqrt(ax2 + ay2)
= 1.217 m/s2
tan(theta) = ax/ay
theta = 81deg
Therefore aT = 1.217 m/s2 [S 81deg E]
(b) For b, I know I would split up DT 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
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?