Parametric acceleration

Calculus Level 1

A particle has position r ⃗ = ( 1 3 t 3 − t , 1 2 t 2 + 3 t − 1 , t 2 ) . \vec{r} = \left(\frac13t^3 - t, \frac12t^2 + 3t-1, t^2\right). What is the magnitude of its acceleration at t = 1 ? t=1?


The answer is 3.

July Thomas by July Thomas , 30, USA

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1 solution

Hana Wehbi
May 18, 2016

To find the acceleration, we need the second derivative of the position vector: r ′ ′ = ( 2 t , 1 , 2 ) r''= (2t,1,2) . At t = 1 t=1 ; r ′ ′ = ( 2 , 1 , 2 ) r''=(2,1,2)

Taking the magnitude: 2 2 + 1 2 + 2 2 = 3 \sqrt {2^{2}+1^{2}+2^{2}}=3

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