Answer to Question #161113 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^2 - 13t + 5. Determine the acceleration of the particle a t = 5.

Answer 1

$x(t) = -t^2 -13t +5 \\ \dfrac{dx(t)}{dt}= -2t-13\\ \dfrac{d^2x(t)}{dt^2} = -2$

But the acceleration of the particle is denoted as $\dfrac{d^2x(t)}{dt^2}$ so at t= 5, we have that $\dfrac{d^2x(t)}{dt^2} = -2$ .So the acceleration of the particle at time t=5secs is $-2ms^{-2}$