Finding the Position of a Spaceship with Given Acceleration and Velocity Vectors

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The discussion focuses on determining the position of a spaceship given its acceleration and initial conditions. The acceleration vector is a(t)=<e^(t), t, sin(2t)>, with initial conditions r(0)=<0,0,0> and v(0)=<1,0,0>. The user initially struggles with finding the correct velocity and position vectors but eventually adjusts the velocity to v(t)=<e^(t), (1/2)t^2, (-1/2)cos(2t)+(1/2)> to satisfy the initial condition. The final position vector is calculated as r(t)=<e^(t)-1, (1/6)t^3, (-1/4)sin(2t)+(1/2)t>, leading to the position at t=pi as r(pi)=<e^(pi)-1, (1/6)pi^3, (pi/2)>. The user seeks confirmation on the accuracy of their solution.
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Finding a position vector...

Homework Statement


A spaceship is traveling with acceleration a(t)=<e^(t) , t , sin2t>. At t=0, the spaceship was a origin r(0)=<0,0,0> and had an initial velocity of v(0)=<1,0,0> Find the position of the ship at t=pi

Homework Equations



uhhh...

The Attempt at a Solution



I figured I'd work backwards with the acceleration vector given to find the velocity and position vectors, then put the t=pi into the position vector I find.

So, v(t)=<e^(t) , (1/2)t^(2) , -(1/2)cos(2t)>
Then r(t)=<e^(t), (1/6)t^(3) , -(1/4)sin(2t)>

But this doesn't seem right to me because when I put in the given t=0 into my v(t), I get <1,0,-(1/2)> not the <1,0,0> like the problem says. I'm approaching this problem wrong, aren't I?
 
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No, you are doing fairly well. What would happen if you added a constant to the velocity?
 


Okay, I have it figured out.

v(t)=<e^(t) , (1/2)t^2 , (-1/2)cos(2t)+(1/2)> So that V(0)=<1,0,0>

Then, I find r(t) which is equal to <e^(t)-1 , (1/6)t^3 , (-1/4)sin(2t)+(1/2)t> So that r(0)=<0,0,0>

Then I plug in pi into my r(t) which comes out to be r(pi)=<e^(pi)-1 , (1/6)pi^3 , (pi/2)> Please tell me this is right. :)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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