Example 30: Car with $x=t^2\left(2-\frac{t}{3}\right)$⁠: time to reach Q and distance PQ

Basics you need

Solution

1. Expand the formula for the distance

   $$x = t^2\left(2 - \frac{t}{3}\right) = 2t^2 - \frac{t^3}{3}.$$
   A sum of powers of $t$ can be differentiated term by term with the power rule. Expanding first avoids the product rule (which would give the same answer with more work).
   Distributive law: $t^2 \cdot 2 = 2t^2$ and $t^2 \cdot \frac{t}{3} = \frac{t^3}{3}$⁠.

2. Differentiate: the velocity

   $$v = \frac{dx}{dt} = 2 \cdot 2t - \frac{3t^2}{3} = 4t - t^2 = t(4 - t).$$

   The unit of $v$ is metres per second, because $x$ is in metres and $t$ in seconds.

   "Stops" is a statement about speed, not about position, so the question is really about $v$⁠. Factorising $4t - t^2$ by taking out the common factor $t$ makes it easy to see where $v$ is zero and what sign it has.
   Power rule: $\frac{d}{dt}(2t^2) = 4t$ and $\frac{d}{dt}\left(\frac{t^3}{3}\right) = \frac{3t^2}{3} = t^2$⁠. Basics: rate of change as a derivative

3. When is the car at rest?

   $$v = 0 \iff t(4 - t) = 0 \iff t = 0 \ \text{ or } \ t = 4.$$

   At $t = 0$ the car is at P, just starting (it starts from rest). So $t = 0$ is not the arrival at Q. The other root, $t = 4$⁠, is the candidate for the stop at Q.

   A product is zero only when one of its factors is zero, so these are the only two instants with $v = 0$⁠. The question tells us the car starts at P at $t = 0$⁠; the zero velocity there is the moment of starting, not of stopping.
   Zero-product rule: $ab = 0$ exactly when $a = 0$ or $b = 0$⁠. Basics: factorising

4. Check that the car keeps moving until $t = 4$

   For $0 < t < 4$⁠: the factor $t$ is positive and the factor $4 - t$ is positive, so

   $$v = t(4 - t) > 0 \quad \text{for } 0 < t < 4.$$

   So after leaving P the car moves forward the whole time and does not stop before $t = 4$⁠. The first stop after the start is at $t = 4$⁠: the car reaches Q after 4 seconds.

   If $v$ became zero somewhere between 0 and 4, the car would have stopped there first, and that point would be Q. The sign check rules this out. It also shows that $x$ is increasing on $[0, 4]$ (Theorem 1 of Section 6.3: $x$ is continuous and $\frac{dx}{dt} > 0$ inside the interval). In fact, because $v$ is strictly positive, the Mean Value Theorem step in the book's proof of Theorem 1 gives $x(t_2) > x(t_1)$ whenever $t_1 < t_2$⁠: $x$ is strictly increasing, so the car never moves backwards.
   Sign of a product: both factors positive gives a positive product. Basics: sign of a product

5. The distance from P to Q

   $$\begin{aligned} x\big|_{t=4} &= 4^2\left(2 - \frac{4}{3}\right) \\ &= 16 \cdot \frac{2}{3} = \frac{32}{3}\ \text{m}. \end{aligned}$$

   So PQ $= \frac{32}{3}$ m $= 10\frac{2}{3}$ m, about 10.67 m.

   $x$ is the distance covered since leaving P. Because the car moved only forward (Step 4), the distance it covered is the same as the straight-line distance between P and Q.
   $2 - \frac{4}{3} = \frac{6}{3} - \frac{4}{3} = \frac{2}{3}$⁠.

6. Why the formula is not used after $t = 4$

   For $t > 4$ the factor $4 - t$ is negative, so the formula would give $v < 0$⁠, and $x$ would fall back to 0 at $t = 6$⁠. A negative "distance covered" rate is impossible for a car that has stopped at Q, so the formula only describes the journey for $0 \le t \le 4$⁠.

   This answers a natural doubt: the formula is a model of one journey. Reading the graph, a student might be tempted to say the car "returns to P at $t = 6$", but the question says the car stops at Q.
   A model is valid only on the range of the situation it describes. Basics: building the function

Every fact used, and where it comes from