The acceleration of a moving object is given by $\frac{d^{2} x}{d t^{2}}=x\left(\frac{d x}{d t}-1\right)$ where x(t)is the position at time t.Which of the following is a system of first order differential equations for position x and velocity v?
A) $v=\frac{d x}{d t}, \frac{d v}{d t}=x(1-v)$
B) $v=\frac{d x}{d t}, \frac{d v}{d t}=v-1$
C) $v=\frac{d x}{d t}, \frac{d v}{d t}=x(v-1)$
D) $v=\frac{d x}{d t}, \frac{d v}{d t}=\frac{x^{2}}{2}(v-1)$

Question

Quizplus · Accepted Answer

Given the acceleration equation, transforming it into a system of first-order differential equations involves expressing the acceleration ($\frac{d^{2} x}{d t^{2}}$) in terms of velocity ($v$) and its derivative ($\frac{d v}{d t}$). The correct transformation is $v=\frac{d x}{d t}$ and $\frac{d v}{d t}=x(v-1)$, which correctly represents the original second-order equation as a system of first-order equations.

The Acceleration of a Moving Object Is Given By d2xdt2=x(dxdt−1)\frac{d^{2} x}{d t^{2}}=x\left(\frac{d x}{d t}-1\right)dt2d2x​=x(dtdx​−1)

The Acceleration of a Moving Object Is Given By $\frac{d^{2} x}{d t^{2}}=x\left(\frac{d x}{d t}-1\right)$