A particle moves in x-y plane so that at any time t its coordinates are x = t^2 - 1, and y = t^4 - 2t^3. What is its acceleration vector at t = 1?

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Answer 1 · 2018-05-12T13:57:43

The answer is #=<2, 0>#

The position is

#{(x=t^2-1),(y=t^4-2t^3):}#

The velocity is the derivative of the position

#{(x'=2t),(y'=4t^3-6t^2):}#

The acceleration is the derivative of the velocity

#{(x''=2),(y''=12t^2-12t):}#

When, #t=1#

The acceleration vector is

#<2, 0>#