Answer to Question #161397 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 at t = 5.

Answer 1

Consider the position function $x(t)=-t^2-13t+5$

Differentiate $x(t)$ with respect to $t$ in order to find the velocity function $v(t)$ as,

$v(t)=\frac{d}{dt}[x(t)]$

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

$=-(2t)-13(1)$

$=-2t-13$

Differentiate $v(t)$ with respect to $t$ in order to find the acceleration function $a(t)$ as,

$a(t)=\frac{d}{dt}[v(t)]$

$=\frac{d}{dt}(-2t-13)$

$=-2(1)-0$

$=-2$

Here, $a(t)=-2$ is a constant function, so the acceleration is constant for $t\geqslant0$ .

Therefore, the acceleration of the particle at $t=5$ is $a(5)=-2$ .