A car passed a point on a course at exactly 12 noon and maintained a speed of 60 km/h. A second car passed the same point 1 hour later, followed the same course, and maintained a speed of 100km/h. When, and after what distance from this point, would the second car have caught up to the first car?
time passed since the second car passed the check point ---- t hours
distance covered by first car = 60 + 60t
distance covered by the 2nd car = 100t
they are equal, take it from there
d1 = d2,
60 * T = 100 *(T-1),
60T = 100T - 100,
T = 2.5 h.
d = 60*T = 60*2.5 = 150 km.
Thank you, Reiny.
To find when the second car catches up to the first car, we can set up an equation based on the information given.
Let's assume that the time it takes for the second car to catch up to the first car is T hours. Since the second car passed the same point 1 hour later, the first car has already been traveling for T + 1 hours.
Since the first car maintained a speed of 60 km/h, the distance it has traveled is 60 * (T + 1) km.
Similarly, since the second car maintained a speed of 100 km/h, the distance it has traveled when it catches up to the first car is 100 * T km.
At the moment both cars meet, the distance traveled by the first car must be equal to the distance traveled by the second car. So we can set up the equation:
60 * (T + 1) = 100 * T
Simplifying the equation:
60T + 60 = 100T
Subtracting 60T from both sides:
60 = 40T
Dividing both sides by 40:
T = 1.5
Therefore, the second car will catch up to the first car after 1.5 hours. To find the distance at which this happens, we can substitute T back into one of the distance equations.
The distance traveled by the second car when it catches up is:
100 * T = 100 * 1.5 = 150 km
So, the second car will catch up to the first car 1.5 hours after passing the point, at a distance of 150 kilometers from that point.