A good name for this would be appreciated

Classical Mechanics Level 2

A particle moved in a straight line so that position x x cm relative to O O at time t t seconds is given by x = t 3 − 6 t 2 + 5 , t ≥ 0 x = t^3 - 6t^2 +5, t \geq 0 . Find the sum of the magnitudes of its initial position, speed and acceleration. If velocity or acceleration point in the minus x x direction, add that magnitude as a negative value to the sum.


The answer is -7.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

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2 solutions

Pranjal Jain
Oct 4, 2014

x = f ( t ) = t 3 − 6 t 2 + 5 x=f(t)=t^{3}-6t^{2}+5 x t = 0 = 5 x_{t=0}=5 v = f ′ ( t ) = 3 t 2 − 12 t v=f'(t)=3t^{2}-12t v t = 0 = 0 v_{t=0}=0 a = f ′ ′ ( t ) = 6 t − 12 a=f''(t)=6t-12 a t = 0 = − 12 a_{t=0}=-12

x + v + a = 5 + 0 + ( − 12 ) = − 7 x+v+a=5+0+(-12)=\boxed{-7}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

te second derivation of distance is acceleration and the first is the speed. it's just to substitute the values for t=0 and sum the results.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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