17. Applications of Differential Equations

d. A Robot Car

Contributed by David Barrow

We will consider a factory which uses a robotic car to move along a linear track on the plant floor to pick up and deliver parts to various locations. The car moves along a straight path on the floor which contains magnetically encoded speed instructions which the car is able to sense. In other words, if we let \(x\) denote the distance along the path from some fixed starting point, then the robot is to move with velocity \(\dfrac{dx}{dt}=v(x)\) with a specific function \(v(x)\). The dependence of \(v\) on \(x\) reflects the fact that the speed instruction is encoded in the floor at position \(x\).

Suppose the speed of a robot car at position \(x\) is \(v(x)=\dfrac{10}{3x^2+1}\,\dfrac{\text{m}}{\text{sec}}\) and it travels from \(x=0\) to \(x=5\) m. Find the total elapsed time.

The position of the robot can be obtained from the initial value problem: \[ \dfrac{dx}{dt}=v(x)=\dfrac{10}{3x^2+1} \qquad \text{with} \qquad x(0)=0 \] The solution is obtained by separation of variables: \[ \int (3x^2+1)\,dx=\int 10\,dt \qquad \Longrightarrow \qquad x^3+x=10t+C \] Using the initial condition, \(x=0\) when \(t=0\), we find \(C=0\). So the solution is \[ x^3+x=10t \] Solving for \(t\) we have: \[ t=\dfrac{x^3+x}{10} \] Since the total elapsed distance is \(x=5\) m, the total elapsed time is: \[ t=\dfrac{5^3+5}{10}=13\,\text{sec} \]