Answer to Question #161398 in Calculus for Johnny Tim

Question

A particle moves along the x-axis so that at time t≥0 its position is given by x(t) = -t^3 + 14t^2 25t. Determine the velocity of the particle at t = 2.

Answer 1

$Given:\;\;Particle's\;position\;is\;given\;by \to\;\;x(t)=-t^3+14t^2+25t$

$To\;find:\;Velocity\;of\;the\;particle\;at\;t=2,\;i.e.,\;x'(2)$

$We\;know\;that\;-\;\;v(t)=\dfrac{d}{dt}(x(t))=\dfrac{d}{dt}(-t^3+14t^2+25t)$

$=-\dfrac{d}{dt}(t^3)+14\dfrac{d}{dt}(t^2)+25\dfrac{d}{dt}(t)$

$=-(3t^2)+14(2t)+25(1)$

$=-3t^2+28t+25$

$\therefore x'(t)=-3t^2+28t+25\;and\;\;x'(2)=-3(2^2)+28(2)+25=-12+81=\mathbf{69}$