Task 16. Points $\boldsymbol{A}$ and $\boldsymbol{B}$ are located on a straight highway running from west to east. Point B is 9 km east of A. A car departs from point A heading east at a speed of 40 km/h. Simultaneously, a motorcycle departs from B in the same direction with a constant acceleration of 32 km/h².

![](https://cdn.mathpix.com/cropped/2024_05_21_07b5a7f6f561b0ff634dg-21.jpg?height=360&width=437&top_left_y=2007&top_left_x=364)

Fig. 21 Determine the maximum distance that can be between the car and the motorcycle during the first two hours of travel.
## S o l u t i o n.

At time $t$, the car is at a distance of $40 t$ km from point $A$, and the motorcycle is at a distance of $\left(16 t^{2}+9\right)$ km.

20

The distance between them is the absolute value of the difference between $16 t^{2}+9$ and $40 t$. Let's denote this distance by $y$:

$$
y=\left|16 t^{2}+9-40 t\right|
$$

We need to find the maximum value of $y$ if $t$ varies in $[0 ; 2]$.

Let's plot the graph of the function $y=$ $=\left|16 t^{2}-40 t+9\right|$ (Fig. 22). If $t$ varies from 0 to 2, then the point on the graph with the greatest ordinate has an abscissa of $\frac{5}{4}$:

$$
y_{\text {max }}=16 \text { (km). }
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_07b5a7f6f561b0ff634dg-22.jpg?height=614&width=539&top_left_y=247&top_left_x=1164)

Fig. 22

Let's trace the graph to see how the distance between the car and the motorcycle changed during the first two hours. At the start of the journey, when $t=0$, the distance between them was 9 km. Over the first 15 minutes, the car was catching up to the motorcycle and at a distance of 1 km from point $B$, it overtook the motorcycle. Then the car started to pull ahead, and the distance between them began to increase. At $t=\frac{5}{4}$ hours, the distance between them reached its maximum value of 16 km. Thereafter, the car continued to lead, but the distance began to decrease as the motorcycle's speed increased and it approached the car. For example, at $t=2$ hours, $y=7$ km, and at $t=\frac{9}{4}$ hours, the motorcycle caught up with the car. If $t>\frac{9}{4}$, the distance between the car and the motorcycle increases again, but this time the motorcycle is in the lead. For instance, at $t=3$ hours, $y=33$ km.