Find the net force with the given position vector

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To find the net force on a 3000N helicopter at t = 5s, the position vector is analyzed, leading to the calculation of the first and second derivatives for acceleration. The acceleration is derived as a = (0.12 m/s^3)(t)i - 0.12 m/s^2k, resulting in a specific value at t = 5s. The mass of the helicopter is calculated using its weight, yielding approximately 305.8 kg. The net force is then determined using F = m*a, resulting in a force of -183.48N. The calculations and thought process presented are correct for determining the net force.
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1.The position vector of a 3000N helicopter is given by:
r=(.02 m/(s^3))(t^3)i + (2.2m/s)tj - (0.06 m/(s^2))(t^2)k
Find the net force on the helicopter at t = 5s.



2. I really have no clue where to begin.



3. I started doing this:
r= (.02m/(s^3))(5s^3)i + (2.2m/s)(5s)j - (.06m/(s^2)*(5s)^2k
r= 2.5mi + 11m + 1.5m
then I drew a 3d graph with each axis labeled and two right triangles but have no clue
where to go from here.
 
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The given problem can be written as

r = a*t^3*i + b*t*j -c*t^2*k

Find d(r/dt and d^2(r)/dt^2. That gives you the acceleration. Find the magnitude of a.

Weight of helicopter is given. Find its mass. Then the force on the helicopter = m*a
 
Thank you so mu rl.bhat.

After the 2nd derivative I get a= (.12m/s^3)(t) - .12m/s^2
= (.6m/s^2)- .12m/s^2= -.6 m/s^2
F=m*a m=F/a m=3000n/9.81m/s^2=305.8kg

m*a=F 305.8 kg * -.6m/s^2=-183.48N Is that thought process correct?
 
After the 2nd derivative I get a= (.12m/s^3)(t) - .12m/s^2

It should be

a= (.12m/s^3)(t)*i - .12m/s^2*k

Find the resultant a.
 
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