The car traveled 240 km between points A and B at a constant speed. Returning back, he covered half the way

| December 7, 2020 | education

The car traveled 240 km between points A and B at a constant speed. Returning back, he covered half the way at the same speed, and then increased the speed by 10 km.h. As a result, the return journey took 24 minutes. less than the path from A to B. How fast did the car go from point A to point B?

Let x be the speed of the car “there” and “back on the first half of the journey”. Then the car passed the rest of the way back at a speed of x + 10. Let us denote S = 240 km – the length of the path. Then the time to travel “there”: t1 = S / x. On the way “back”: t2 = 0.5S / x + 0.5S / (x + 10) = t1 – 0.4 (24min = 0.4h). Substituting the t1 value: 0.5S / x + 0.5S / (x + 10) = S / x – 0.4. Solving the equation, we find x: X1 = 50, X2 = -60. The speed cannot be negative => discard the second option. Remains x = 50km / h.
Answer: 50 km / h.



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