Answer:
At t=4.82 s, the boat is moving at 3.464 m/s.
At t=4.82 s, the boat is 13.112 m from the dock.
Explanation:
The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.
In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:
vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s
Now, we add this velocity to the velocity of the water in the i direction:
vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s
Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.
For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.
The speed in the i'th direction, for all times, is given by:
(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:
di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m
The speed in j'th direction, for all times, is given by:
2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:
dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m
Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.
so if we double the charge on 
and double the distance to 
we plug these into the equation to find
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if I can remember my physics correctly.