Physics Help! The velocity of a 3.00kg particle is given by v=(8.00ti + 3.00t^2j) m/sFull question below?

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The velocity of a 3 kg particle is given by v = (8.00ti + 3.00t^2j)m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 43 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?

I have seen the other answers on Yahoo Answers and they do not come out correct.

Please help.  Very frustrated here.  

I derived vel to get accel = (8.00 + 6.00t)

I tried solving for t at the instant the magnitude is 35.0N by
35=3kg (8.00 + 6.00t) ---> 35=24i+18tj
11=18t ---> t=0.611

I used t back into v= and a= equations but get stuck here.  Nothing I do will get me the answer

Anonymous · Accepted Answer

There are two mistakes in your solution.
1. The instant magnitude of force is given as 43N where as you have taken it as 35N.
2.You have equated a scalar quantity to a vector quantity. It is absurd.
Find the solution in the following manner:
step 1: find inst acceleration(a) = F/m = 43N/3kg=14.33m/s^2
step 2: Equate 14.33m/^2 to the magnitude of (8i+6tj)
=> t = 1.98s
step 3: to get the direction of force when it is 43N, substitue the value of t in 8i+6tj
=> 8i+6x1.98j
=>8i+11.89j
=> theta = arctan(11.89/8)
=> theta=56.07 degrees
b) The particle will move in 2 dimensions, in the direction of applied force.

Lord Cracker Jack · Answer

I use to be good at math....until the alphabet got involved :/

Physics Help! The velocity of a 3.00kg particle is given by v=(8.00ti + 3.00t^2j) m/sFull question below?

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