Exercise 4
The purpose of this exercise is to study the stopping of a cart on a ride, from the moment it enters the braking zone at the end of the course. We denote $t$ the elapsed time, expressed in seconds, from the moment the cart enters the braking zone. We model the distance travelled by the cart in the braking zone, expressed in metres, as a function of $t$, using a function denoted $d$ defined on $[0; +\infty[$. We thus have $d(0) = 0$. Furthermore, we admit that this function $d$ is differentiable on its domain of definition. We denote $d'$ its derivative function.
Part AIn the figure (Fig. 2), we have drawn in an orthonormal coordinate system:
- the representative curve $\mathscr{C}_d$ of the function $d$;
- the tangent $T$ to the curve $\mathscr{C}_d$ at point A with abscissa 4.7;
- the asymptote $\Delta$ to $\mathscr{C}_d$ at $+\infty$.
In this part, no justification is expected. With the precision that the graph allows, answer the questions below. According to this model:
- After how much time will the cart have travelled 15 m in the braking zone?
- What minimum length must be provided for the braking zone?
- What is the value of $d'(4{,}7)$? Interpret this result in the context of the exercise.
Part BWe recall that $t$ denotes the elapsed time, in seconds, from the moment the cart enters the braking zone. We model the instantaneous velocity of the cart, in metres per second ($\mathrm{m.s^{-1}}$), as a function of $t$, by a function $v$ defined on $[0; +\infty[$. We admit that:
- the function $v$ is differentiable on its domain of definition, and we denote $v'$ its derivative function;
- the function $v$ is a solution of the differential equation $$(E): \quad y' + 0{,}6\, y = \mathrm{e}^{-0{,}6t},$$ where $y$ is an unknown function and $y'$ is the derivative function of $y$.
We further specify that, upon arrival in the braking zone, the velocity of the cart is equal to $12\,\mathrm{m.s^{-1}}$, that is $v(0) = 12$.
- [a.] We consider the differential equation $$(E'): \quad y' + 0{,}6\, y = 0$$ Determine the solutions of the differential equation $(E')$ on $[0; +\infty[$.
- [b.] Let $g$ be the function defined on $[0; +\infty[$ by $g(t) = t\,\mathrm{e}^{-0{,}6t}$. Verify that the function $g$ is a solution of the differential equation $(E)$.
- [c.] Deduce the solutions of the differential equation $(E)$ on $[0; +\infty[$.
- [d.] Deduce that for every real $t$ belonging to the interval $[0; +\infty[$, we have: $$v(t) = (12 + t)\,\mathrm{e}^{-0{,}6t}$$
- In this question, we study the function $v$ on $[0; +\infty[$.
- [a.] Show that for every real $t \in [0; +\infty[$, $v'(t) = (-6{,}2 - 0{,}6t)\,\mathrm{e}^{-0{,}6t}$.
- [b.] By admitting that: $$v(t) = 12\,\mathrm{e}^{-0{,}6t} + \frac{1}{0{,}6} \times \frac{0{,}6t}{\mathrm{e}^{0{,}6t}}$$ determine the limit of $v$ at $+\infty$.
- [c.] Study the direction of variation of the function $v$ and draw up its complete variation table. Justify.
- [d.] Show that the equation $v(t) = 1$ has a unique solution $\alpha$, of which you will give an approximate value to the nearest tenth.
- When the velocity of the cart is less than or equal to 1 metre per second, a mechanical system is triggered allowing its complete stopping. Determine after how much time this system comes into action. Justify.
Part CWe recall that for every real $t$ belonging to the interval $[0; +\infty[$: $$v(t) = (12 + t)\,\mathrm{e}^{-0{,}6t}.$$ We admit that for every real $t$ in the interval $[0; +\infty[$: $$d(t) = \int_0^t v(x)\,\mathrm{d}x$$
- Using integration by parts, show that the distance travelled by the cart between times 0 and $t$ is given by: $$d(t) = \mathrm{e}^{-0{,}6t}\left(-\frac{5}{3}t - \frac{205}{9}\right) + \frac{205}{9}$$
- We recall that the stopping device is triggered when the velocity of the cart is less than or equal to 1 metre per second. Determine, according to this model, an approximate value to the nearest hundredth of the distance travelled by the cart in the braking zone before the triggering of this device.