Question

question 2: a cnc machine head moves with an acceleration of a = - 0.3v mm/s². when t = 0, the head is at the origin of the machine, but has an initial velocity of 40 mm/s. what is the displacement (in mm) after 3.5s? what is the acceleration at this point?

Explanation:

Step1: Recall the acceleration - velocity relationship

We know that $a=\frac{dv}{dt}=- 0.3v$. Rearranging this differential equation gives $\frac{dv}{v}=-0.3dt$.

Step2: Integrate both sides

Integrating $\int_{v_0}^{v}\frac{dv}{v}=\int_{0}^{t}-0.3dt$. The left - hand side integral is $\ln v-\ln v_0=\ln\frac{v}{v_0}$, and the right - hand side is $-0.3t$. So, $\ln\frac{v}{v_0}=-0.3t$, and $v = v_0e^{-0.3t}$.

Step3: Recall the velocity - displacement relationship

Since $v=\frac{dx}{dt}=v_0e^{-0.3t}$, we can integrate to find the displacement. $\int_{0}^{x}dx=\int_{0}^{t}v_0e^{-0.3t}dt$.

Step4: Integrate the right - hand side

Let $u=-0.3t$, then $dt=-\frac{1}{0.3}du$. $\int_{0}^{t}v_0e^{-0.3t}dt=v_0\int_{0}^{-0.3t}-\frac{1}{0.3}e^{u}du=\frac{v_0}{0.3}(1 - e^{-0.3t})$.
Given $v_0 = 40$ mm/s and $t = 3.5$ s.
$x=\frac{40}{0.3}(1 - e^{-0.3\times3.5})$
$x=\frac{40}{0.3}(1 - e^{-1.05})$
$x=\frac{40}{0.3}(1 - 0.3499)$
$x=\frac{40}{0.3}\times0.6501\approx86.68$ mm

To find the acceleration at $t = 3.5$ s, first find the velocity at $t = 3.5$ s.
$v=v_0e^{-0.3t}=40e^{-0.3\times3.5}=40\times0.3499 = 13.996$ mm/s
Then, since $a=-0.3v$, $a=-0.3\times13.996\approx - 4.20$ mm/s²

Answer:

The displacement is approximately $86.68$ mm and the acceleration is approximately $-4.20$ mm/s².