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Homework problem

  • #1
I wanted to make sure I did this problem correctly. The problem is:

A particle moves along a straight line and its position at time t is given by s(t)=2t^3 - 21t^2 + 60t where s is measured in feet and t in seconds.

Find the velocity of the particle when t=0.

I took the derivative of s(t) and got 6t^2 - 42t + 60, and then substituted 0 in for t. I got 60ft/sec for the answer. Is this correct??
 

Answers and Replies

  • #2
4
0
[tex]\frac{ds}{dt}=v(t)[/tex]

[tex]\frac{dv}{dt}=a(t)[/tex]

Where [tex]s(t)[/tex] is the displacement vector, [tex]v(t)[/tex] is the velocity vector and
[tex]a(t)[/tex] is the acceleration vector.

If you look at it in any given point, so I'm guessing that it's correct.
 

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