A cyclist departs from point A towards point B, which are 120 km apart. After some time, a car departs from point A in the same direction. The car reaches point B, stops for 3 hours, and then drives back to A at the same speed. The graph shows the movement of the cyclist (curve 1) and the car (curve 2) only on the way from A to B. The horizontal axis represents time, and the vertical axis represents the distance from point A. Find the distance from point A at which the car overtook the cyclist.

Solution:

To find the distance at which the car overtook the cyclist, we need to find the point of intersection of the two graphs (curve 1 and curve 2). From the graph, we can see that the two curves intersect at approximately the following coordinates:

• Time (horizontal axis): Around 13-14 hours.
• Distance from point A (vertical axis): Around 70-80 km.

Let's estimate the intersection point more precisely from the graph:

• Curve 1 (cyclist) passes through points like (0, 0), (approx 13, 70), (approx 15, 80), (approx 17, 100), (approx 19, 110), (24, 120).
• Curve 2 (car) passes through points like (approx 10, 0), (approx 13, 70).

The intersection appears to be around the time when the car catches up to the cyclist. Observing the graph, the intersection point seems to occur roughly when the car's path (curve 2) crosses the cyclist's path (curve 1).

Looking closely at the grid:

• The car's graph (curve 2) starts at approximately time 10 and reaches a distance of around 70-80 km by time 13.
• The cyclist's graph (curve 1) shows that at time 13, the distance is also around 70-80 km.
• The exact point of intersection is where the distance from point A is the same for both the cyclist and the car. This happens when curve 2 crosses curve 1.
• By visual inspection of the graph, the intersection point appears to be at a distance of approximately 75 km from point A, which occurs at approximately 13.5 hours.

To confirm this, let's estimate the speeds. The cyclist travels 120 km in 24 hours, so the average speed of the cyclist is 120/24 = 5 km/h. However, the graph shows the cyclist's speed is not constant. For the first ~10 hours, the cyclist travels ~50 km, so the speed is 5 km/h. From hour 10 to hour 15, the cyclist travels from 50 km to 80 km (30 km in 5 hours), speed is 6 km/h. From hour 15 to hour 19, travels from 80 km to 110 km (30 km in 4 hours), speed is 7.5 km/h. From hour 19 to 24, travels from 110 km to 120 km (10 km in 5 hours), speed is 2 km/h.

The car starts at time 10. Let's assume it travels at a constant speed to point B (120 km away). Let's approximate the time the car reaches point B from the graph. It looks like it reaches around 14.5-15 hours. If it takes 4.5 hours to travel 120 km, the speed is 120 / 4.5 = 26.67 km/h. However, the car stops for 3 hours. The question states the car's graph is only on the way FROM A to B. The car departs at some time after the cyclist, let's say at time $$t_c$$. The cyclist departs at time 9. Let's assume the car departs at time 10 as suggested by graph 2 starting at 10. So, at time $$t$$, the distance of the car is $$d_c(t) = v_c imes (t - 10)$$ for $$t imes 10$$. The cyclist's distance at time $$t$$ is $$d_{cy}(t)$$. We are looking for $$t$$ such that $$d_c(t) = d_{cy}(t)$$.

From the graph, at time 13, the cyclist is at approximately 70 km. The car is also at approximately 70 km at time 13. This implies the car overtook the cyclist at this point. Let's try to be more precise. The car's path (curve 2) seems to cross the cyclist's path (curve 1) at approximately $$t=13.5$$ hours. At this time, the distance is approximately 75 km. Let's check if this is consistent. If the car travels 75 km in 3.5 hours (from 10 to 13.5), its speed is $$75 / 3.5 imes 7 = 150/7 imes 7 = 150$$ km in 7 hours. This doesn't seem right. Let's re-examine the graph. The point where curve 2 intersects curve 1 is crucial.

Let's assume the intersection is precisely at the grid lines. The intersection occurs between 13 and 14 on the time axis and between 70 and 80 on the distance axis. It looks like it is closer to 13.5 hours and around 75 km.

Let's try to be more accurate by looking at the segments. For the cyclist, from $$t=10$$ to $$t=15$$, the distance increases from 50 to 80. This is a rate of $$(80-50)/(15-10) = 30/5 = 6$$ km/h. For the car, it starts at $$t=10$$. Let's assume it reaches B at $$t=15$$. Then its speed is $$120/(15-10) = 120/5 = 24$$ km/h. If the car's speed is 24 km/h and it starts at $$t=10$$, its distance at time $$t$$ is $$d_c(t) = 24(t-10)$$. For the cyclist, let's approximate the segment from $$t=10$$ to $$t=15$$ as linear. The equation for the cyclist's path in this interval is $$d_{cy}(t) = 50 + 6(t-10)$$. We want to find $$t$$ where $$d_c(t) = d_{cy}(t)$$.

$$24(t-10) = 50 + 6(t-10)$$

$$24t - 240 = 50 + 6t - 60$$

$$24t - 240 = 6t - 10$$

$$18t = 230$$

$$t = 230/18 = 115/9 imes 60 = 12.78$$ hours.

At $$t=12.78$$ hours, the distance is $$d_c(12.78) = 24(12.78-10) = 24 imes 2.78 = 66.72$$ km. This is not consistent with the graph. The graph shows the intersection at a higher distance.

Let's re-examine the graph carefully. The intersection point is very close to the grid line representing 13.5 hours and the grid line representing 75 km. Let's assume the intersection is at $$t = 13.5$$ hours. The distance covered by the cyclist up to this time is approximately 75 km (midpoint between 70 and 80). If the car starts at 10 hours and reaches 75 km at 13.5 hours, its travel time is 3.5 hours. Its speed would be $$75 / 3.5 = 750/35 = 150/7 imes 7 imes 5 = 21.43$$ km/h. This speed is plausible.

Let's check if the cyclist's distance is approximately 75 km at 13.5 hours. From $$t=10$$ to $$t=15$$, the cyclist moves from 50 km to 80 km. So at $$t=13.5$$, which is halfway between 10 and 15 (relative to the segment), the distance should be around the midpoint of 50 and 80, which is 65 km. However, the graph shows the intersection is higher. The segment from $$t=10$$ to $$t=15$$ for the cyclist is not a straight line, it bends upwards, meaning the speed increases. The segment from 10 to 15 for the cyclist looks like it goes from (10, 50) to (12, 60) to (13.5, 75) to (15, 80). If we assume the intersection is at (13.5, 75), then the car's speed is indeed $$75/3.5 imes 7 = 21.43$$ km/h. This seems to be the most reasonable interpretation of the graph.

Therefore, the car overtook the cyclist at a distance of approximately 75 km from point A.

Graph Visualization

Note: The above graph is an approximation based on the visual data from the provided image. The cyclist's path is approximated as piecewise linear with increasing speed, and the car's path is assumed to be linear until point B. The intersection point visually appears to be around (13.5, 75).

Answer: 75